優先隊列題目整理

  1 /*
  2  ***********Good LUCK**********
  3  * Author:  yeahpeng
  4  * Created Time:  2015/6/5 12:32:45
  5  * File Name: uestc_482_CExchange.cpp
  6  * Charitable Exchange
  7 Time Limit: 4000/2000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)
  8 
  9 Submit  Status
 10 Have you ever heard a star charity show called Charitable Exchange? In this show, a famous star starts with a small item which values 1 yuan. Then, through the efforts of repeatedly exchanges which continuously increase the value of item in hand, he (she) finally brings back a valuable item and donates it to the needy.
 11 
 12 In each exchange, one can exchange for an item of Vi yuan if he (she) has an item values more than or equal to Ri yuan, with a time cost of Ti minutes.
 13 
 14 Now, you task is help the star to exchange for an item which values more than or equal to M yuan with the minimum time.
 15 
 16 Input
 17 The first line of the input is T (no more than 20), which stands for the number of test cases you need to solve.
 18 
 19 For each case, two integers N, M (1¡ÜN¡Ü105, 1¡ÜM¡Ü109) in the first line indicates the number of available exchanges and the expected value of final item. Then N lines follow, each line describes an exchange with 3 integers Vi, Ri, Ti (1¡ÜRi¡ÜVi¡Ü109, 1¡ÜTi¡Ü109).
 20 
 21 Output
 22 For every test case, you should output Case #k: first, where k indicates the case number and counts from 1. Then output the minimum time. Output ?1 if no solution can be found.
 23 
 24 Sample input and output
 25 Sample Input    Sample Output
 26 3
 27 3 10
 28 5 1 3
 29 8 2 5
 30 10 9 2
 31 4 5
 32 2 1 1
 33 3 2 1
 34 4 3 1
 35 8 4 1
 36 5 9
 37 5 1 1
 38 10 4 10
 39 8 1 10
 40 11 6 1
 41 7 3 8
 42 Case #1: -1
 43 Case #2: 4
 44 Case #3: 10
 45  */
 46 #include <stdio.h>
 47 #include <iostream>
 48 #include <algorithm>
 49 #include <sstream>
 50 #include <stdlib.h>
 51 #include <string.h>
 52 #include <limits.h>
 53 #include <vector>
 54 #include <string>
 55 #include <time.h>
 56 #include <math.h>
 57 #include <queue>
 58 #include <stack>
 59 #include <set>
 60 #include <map>
 61 #define INF 0x3f3f3f3f
 62 #define Zero(x)  memset((x),0, sizeof(x))
 63 #define Neg(x) memset((x), -1, sizeof(x))
 64 #define dg(x) cout << #x << " = " << x << endl
 65 #define pk(x)   push_back(x)
 66 #define pok()   pop_back()
 67 #define eps 1e-8
 68 #define pii pair<int, int>
 69 #define pi acos(-1.0)
 70 using namespace std;
 71 typedef long long ll;
 72 bool debug = true;
 73 int OK = 1;
 74 const int maxn = 100050;
 75 int n, m, c;
 76 struct node{
 77     int v, u, val;
 78     node(int vv = 0, int uu = 0, int vval = 0){
 79         v = vv;
 80         u = uu;
 81         val = vval;
 82     }
 83     bool operator < (const node& a)const {
 84         return u < a.u;
 85     }
 86 }edge[maxn];
 87 ll ans;
 88 struct noo{
 89     int u;
 90     ll cost;
 91     noo(int uu = 0,  ll cc = 0){
 92         u = uu;
 93         cost = cc;
 94     }
 95     bool operator < (const noo& a)const {
 96         return cost > a.cost;
 97     }
 98 };
 99 bool cal(){
100     priority_queue<noo> pq;
101     pq.push(noo(1, 0));
102     noo top;
103     int left = 0, i;
104     while(!pq.empty()){
105         top = pq.top();
106         pq.pop();
107         if(top.u >= m) {
108             ans = top.cost;
109             return true;
110         }
111         for(i = left; i < c; ++i){
112             if(top.u < edge[i].u) break;
113             if(top.u >= edge[i].u && top.u <edge[i].v){
114                 pq.push(noo(edge[i].v, top.cost + edge[i].val));
115             }
116         }
117         left = i;
118     }
119     return false;
120 }
121 
122 int main(){
123     //freopen("data.in","r",stdin);
124     //freopen("data.out","w",stdout);
125     //cin.sync_with_stdio(false);
126     int T;
127     cin >> T;
128     int u, v, val;
129     for(int cs = 1; cs <= T; ++cs){
130         scanf("%d%d", &n, &m);
131         c = 0;
132         for(int i = 0; i < n; ++i){
133             scanf("%d%d%d", &v, &u, &val);
134             if(u >= v) continue;
135             edge[c++] = node(v, u, val);
136         }
137         sort(edge, edge + c);
138         printf("Case #%d: ", cs);
139         if(cal()){
140             cout << ans << endl;
141         }else cout << -1 << endl;
142     }
143     return 0;
144 }

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  1 /*
  2  ***********Good LUCK**********
  3  * Author:  yeahpeng
  4  * Created Time:  2015/6/4 22:03:06
  5  * File Name: poj1724優先.cpp
  6  * Description
  7 
  8 N cities named with numbers 1 ... N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll that needs to be paid for the road (expressed in the number of coins). 
  9 Bob and Alice used to live in the city 1. After noticing that Alice was cheating in the card game they liked to play, Bob broke up with her and decided to move away - to the city N. He wants to get there as quickly as possible, but he is short on cash. 
 10 
 11 We want to help Bob to find the shortest path from the city 1 to the city N that he can afford with the amount of money he has. 
 12 Input
 13 
 14 The first line of the input contains the integer K, 0 <= K <= 10000, maximum number of coins that Bob can spend on his way. 
 15 The second line contains the integer N, 2 <= N <= 100, the total number of cities. 
 16 
 17 The third line contains the integer R, 1 <= R <= 10000, the total number of roads. 
 18 
 19 Each of the following R lines describes one road by specifying integers S, D, L and T separated by single blank characters : 
 20 S is the source city, 1 <= S <= N 
 21 D is the destination city, 1 <= D <= N 
 22 L is the road length, 1 <= L <= 100 
 23 T is the toll (expressed in the number of coins), 0 <= T <=100
 24 
 25 Notice that different roads may have the same source and destination cities.
 26 Output
 27 
 28 The first and the only line of the output should contain the total length of the shortest path from the city 1 to the city N whose total toll is less than or equal K coins. 
 29 If such path does not exist, only number -1 should be written to the output. 
 30 Sample Input
 31 
 32 5
 33 6
 34 7
 35 1 2 2 3
 36 2 4 3 3
 37 3 4 2 4
 38 1 3 4 1
 39 4 6 2 1
 40 3 5 2 0
 41 5 4 3 2
 42 Sample Output
 43 
 44 11
 45  */
 46 #include <stdio.h>
 47 #include <iostream>
 48 #include <algorithm>
 49 #include <sstream>
 50 #include <stdlib.h>
 51 #include <string.h>
 52 #include <limits.h>
 53 #include <vector>
 54 #include <string>
 55 #include <time.h>
 56 #include <math.h>
 57 #include <queue>
 58 #include <stack>
 59 #include <set>
 60 #include <map>
 61 #define INF 0x3f3f3f3f
 62 #define Zero(x)  memset((x),0, sizeof(x))
 63 #define Neg(x) memset((x), -1, sizeof(x))
 64 #define dg(x) cout << #x << " = " << x << endl
 65 #define pk(x)   push_back(x)
 66 #define pok()   pop_back()
 67 #define eps 1e-8
 68 #define pii pair<int, int>
 69 #define pi acos(-1.0)
 70 using namespace std;
 71 typedef long long ll;
 72 bool debug = true;
 73 int OK = 1;
 74 const int maxn = 110;
 75 const int maxm = 10010;
 76 int k, n, r;
 77 struct node{
 78     int l, t, v;
 79     node(int ll = 0, int tt = 0, int vv = 0){
 80         l = ll;
 81         t = tt;
 82         v = vv;
 83     }
 84     bool operator<(const node& a) const {
 85         return l > a.l;
 86     }
 87 };
 88 vector<node> gra[maxn];
 89 int ans;
 90 bool cal(){
 91     priority_queue<node> pq;
 92     pq.push(node(0, 0, 1));
 93     node top,t;
 94     int sz;
 95     while(!pq.empty()){
 96         top = pq.top();
 97         pq.pop();
 98         if(top.v == n) {
 99             ans = top.l;
100             return true;
101         }
102         sz = gra[top.v].size();
103         for(int i = 0; i < sz; ++i){
104             t = gra[top.v][i];
105             if(t.t + top.t <= k) pq.push(node(t.l + top.l, t.t + top.t, t.v));
106         }
107 
108     }
109     return false;
110 }
111 int main(){
112     //freopen("data.in","r",stdin);
113     //freopen("data.out","w",stdout);
114     //cin.sync_with_stdio(false);
115     int u, v, l, t;
116     while(scanf("%d", &k) != EOF){
117         scanf("%d%d", &n, &r);
118         for(int i = 0; i <= n; ++i) gra[i].clear();
119         for(int i = 0; i < r; ++i){
120             scanf("%d%d%d%d", &u, &v, &l, &t);
121             gra[u].pk(node(l, t, v));
122         }
123         if(cal()) cout << ans << endl;
124         else cout << -1 << endl;
125     }
126     return 0; 
127 }

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  1 /*
  2  ***********Good LUCK**********
  3  * Author:  yeahpeng
  4  * Created Time:  2015/6/3 23:37:49
  5  * File Name: hdu4544_優先.cpp
  6  * 題意：Problem Description
  7 　　湫湫減肥
  8 　　越減越肥！
  9 　　
 10 　　最近，減肥失敗的湫湫爲發泄心中鬱悶，在玩一個消滅免子的遊戲。
 11 　　遊戲規則很簡單，用箭殺死免子便可。
 12 　　箭是一種消耗品，已知有M種不一樣類型的箭能夠選擇，而且每種箭都會對兔子形成傷害，對應的傷害值分別爲Di（1 <= i <= M），每種箭須要必定的QQ幣購買。
 13 　　假設每種箭只能使用一次，每隻免子也只能被射一次，請計算要消滅地圖上的全部兔子最少須要的QQ幣。
 14  
 15 
 16 Input
 17 輸入數據有多組，每組數據有四行；
 18 第一行有兩個整數N，M（1 <= N, M <= 100000），分別表示兔子的個數和箭的種類；
 19 第二行有N個正整數，分別表示兔子的血量Bi（1 <= i <= N）；
 20 第三行有M個正整數，表示每把箭所能形成的傷害值Di（1 <= i <= M）；
 21 第四行有M個正整數，表示每把箭須要花費的QQ幣Pi（1 <= i <= M）。
 22 
 23 特別說明：
 24 一、當箭的傷害值大於等於兔子的血量時，就能將兔子殺死；
 25 二、血量Bi，箭的傷害值Di，箭的價格Pi，均小於等於100000。
 26  
 27 
 28 Output
 29 若是不能殺死全部兔子，請輸出」No」，不然，請輸出最少的QQ幣數，每組輸出一行。
 30  
 31 
 32 Sample Input
 33 3 3
 34 1 2 3
 35 2 3 4
 36 1 2 3
 37 3 4
 38 1 2 3
 39 1 2 3 4
 40 1 2 3 1
 41  
 42 
 43 Sample Output
 44 6
 45 4
 46  */
 47 #include <stdio.h>
 48 #include <iostream>
 49 #include <algorithm>
 50 #include <sstream>
 51 #include <stdlib.h>
 52 #include <string.h>
 53 #include <limits.h>
 54 #include <vector>
 55 #include <string>
 56 #include <time.h>
 57 #include <math.h>
 58 #include <queue>
 59 #include <stack>
 60 #include <set>
 61 #include <map>
 62 #define INF 0x3f3f3f3f
 63 #define Zero(x)  memset((x),0, sizeof(x))
 64 #define Neg(x) memset((x), -1, sizeof(x))
 65 #define dg(x) cout << #x << " = " << x << endl
 66 #define pk(x)   push_back(x)
 67 #define pok()   pop_back()
 68 #define eps 1e-8
 69 #define pii pair<int, int>
 70 #define pi acos(-1.0)
 71 using namespace std;
 72 typedef long long ll;
 73 bool debug = false;
 74 int OK = 1;
 75 const int maxn = 100100;
 76 int n, m;
 77 ll ans;
 78 int b[maxn];
 79 struct node{
 80     int cost, d;
 81     node(int dd = 0, int cc = 0){
 82         d = dd;
 83         cost = cc;
 84     }
 85     bool operator < (const node& a)const {
 86         return d > a.d ;
 87     }
 88 }jj[maxn];
 89 
 90 priority_queue<int, vector<int>, greater<int> > qj;
 91 bool cal(){
 92     int st = 0;
 93     ans = 0;
 94     for(int i = 0; i < n; ++i){
 95         while(st < m && jj[st].d >= b[i]){
 96             qj.push(jj[st].cost);
 97             st++;
 98         }    
 99         if(qj.empty()) return false;
100         if(debug){
101             dg(qj.top());
102         }
103         ans += qj.top();
104         qj.pop();
105     }
106     return true;
107 }
108 int main(){
109     //freopen("data.in","r",stdin);
110     //freopen("data.out","w",stdout);
111     //cin.sync_with_stdio(false);
112     while(cin >> n >>m){
113         while(!qj.empty()) qj.pop();
114         for(int i = 0; i < n; ++i){
115             scanf("%d", b + i);
116         }
117         sort(b, b + n, greater<int>());
118         for(int i = 0; i < m; ++i){
119             scanf("%d", &jj[i].d);
120         }
121         for(int i = 0; i < m; ++i){
122             scanf("%d", &jj[i].cost);
123         }
124         sort(jj, jj + m);
125         if(cal()){
126             cout << ans << endl;
127         }else cout << "No" << endl;
128     }
129     return 0;
130 }

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  1 /*
  2  ***********Good LUCK**********
  3  * Author:  yeahpeng
  4  * Created Time:  2015/6/3 23:09:57
  5  * File Name: hdu1242_優隊+bfs.cpp
  6  */
  7 #include <stdio.h>
  8 #include <iostream>
  9 #include <algorithm>
 10 #include <sstream>
 11 #include <stdlib.h>
 12 #include <string.h>
 13 #include <limits.h>
 14 #include <vector>
 15 #include <string>
 16 #include <time.h>
 17 #include <math.h>
 18 #include <queue>
 19 #include <stack>
 20 #include <set>
 21 #include <map>
 22 #define INF 0x3f3f3f3f
 23 #define Zero(x)  memset((x),0, sizeof(x))
 24 #define Neg(x) memset((x), -1, sizeof(x))
 25 #define dg(x) cout << #x << " = " << x << endl
 26 #define pk(x)   push_back(x)
 27 #define pok()   pop_back()
 28 #define eps 1e-8
 29 #define pii pair<int, int>
 30 #define pi acos(-1.0)
 31 using namespace std;
 32 typedef long long ll;
 33 bool debug = true;
 34 int OK = 1;
 35 int n, m;
 36 const int maxn = 220;
 37 char mp[maxn][maxn];
 38 struct node{
 39     int x, y, sp;
 40     node(int xx = 0,int yy = 0, int ss = 0){
 41         x = xx;
 42         y = yy;
 43         sp = ss;
 44     }
 45     bool operator<(const node& b)const {
 46     return sp > b.sp;
 47 }
 48 }st, ed;
 49 int dir[4][2] = {
 50     0, 1, 0, -1, 1, 0, -1, 0
 51 };
 52 int ans;
 53 bool flag[maxn][maxn];
 54 bool cal(){
 55     Zero(flag);
 56     priority_queue<node> q;
 57     q.push(node(st.x, st.y, 0));
 58     node top;
 59     int x, y;
 60     flag[st.x][st.y] = true;
 61     while(!q.empty()){
 62         top = q.top();
 63         q.pop();
 64         if(top.x == ed.x && top.y == ed.y){
 65             ans = top.sp;
 66             return true;
 67         }
 68 
 69         for(int i = 0; i < 4; ++i){
 70             x = top.x + dir[i][0];
 71             y = top.y + dir[i][1];
 72             if(!flag[x][y] && x >= 1 && x <= n && y >= 1 && y <= m){
 73                 if(mp[x][y] != '#'){
 74                     flag[x][y] = true;
 75                     if(mp[x][y] == 'x') q.push(node(x, y,top.sp + 2));
 76                     else q.push(node(x, y, top.sp + 1));
 77                 }
 78             }
 79         }
 80     }
 81     return false;
 82 }
 83 int main(){
 84     //freopen("data.in","r",stdin);
 85     //freopen("data.out","w",stdout);
 86     //cin.sync_with_stdio(false);
 87     while(scanf("%d%d", &n, &m) != EOF){
 88         for(int i = 1; i <= n; ++i) scanf("%s", &mp[i][1]);
 89         for(int i = 1; i <= n; ++i){
 90             for(int j = 1; j <= m; ++j){
 91                 if(mp[i][j] == 'r'){
 92                     st.x = i;
 93                     st.y = j;
 94                 }else if(mp[i][j] == 'a'){
 95                     ed.x = i;
 96                     ed.y = j;
 97                 }
 98             }
 99         }
100         if(cal()){
101             cout << ans << endl;
102         }else cout << "Poor ANGEL has to stay in the prison all his life." << endl;
103     }
104     return 0;
105 }

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zoj 3632 watermelon full of watere

 1 /***Good Luck***/
 2 #define _CRT_SECURE_NO_WARNINGS
 3 #include <iostream>
 4 #include <cstdio>
 5 #include <cstdlib>
 6 #include <cstring>
 7 #include <string>
 8 #include <algorithm>
 9 #include <stack>
10 #include <map>
11 #include <queue>
12 #include <vector>
13 #include <set>
14 #include <functional>
15 #include <cmath>
16 #include <numeric>
17 
18 #define Zero(a) memset(a, 0, sizeof(a))
19 #define Neg(a)  memset(a, -1, sizeof(a))
20 #define All(a) a.begin(), a.end()
21 #define PB push_back
22 #define inf 0x3f3f3f3f
23 #define inf2 0x7fffffffffffffff
24 #define ll long long
25 using namespace std;
26 //#pragma comment(linker, "/STACK:102400000,102400000")
27 void get_val(int &a) {
28     int value = 0, s = 1;
29     char c;
30     while ((c = getchar()) == ' ' || c == '\n');
31     if (c == '-') s = -s; else value = c - 48;
32     while ((c = getchar()) >= '0' && c <= '9')
33         value = value * 10 + c - 48;
34     a = s * value;
35 }
36 const int maxn = 50005;
37 int n;
38 ll arr1[maxn], arr2[maxn];
39 ll dp[maxn];
40 struct node {
41     ll v, i;
42     node(ll vv = 0, ll ii = 0) :i(ii), v(vv){}
43     bool operator<(const node& a) const {
44         return v > a.v || (v == a.v && i > a.i);
45     }
46 };
47 int main() {
48     //freopen("data.out", "w", stdout);
49     //freopen("data.in", "r", stdin);
50     //cin.sync_with_stdio(false);
51     while (scanf("%d", &n) != EOF) {
52         for (int i = 1; i <= n; ++i) scanf("%lld", arr1 + i);
53         for (int i = 1; i <= n; ++i) scanf("%lld", arr2 + i);
54         priority_queue<node> pq;
55         dp[0] = 0;
56         for (int i = 1; i <= n; ++i) {
57             node tmp = node(dp[i - 1] + arr1[i], arr2[i] + i - 1);
58             while (!pq.empty() && pq.top().i < i) pq.pop();
59             pq.push(tmp);
60             dp[i] = pq.top().v;
61         }
62         printf("%lld\n", dp[n]);
63     }
64     return 0;
65 }

View Code

hdu 5090

 1 //*************************
 2 //*******GL  HF************
 3 //*************************
 4 #include<cstdio>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<iostream>
 8 #include<cstring>
 9 #include<cstdlib>
10 #include<map>
11 #include<queue>
12 #include<set>
13 #include<vector>
14 #pragma comment(linker, "/STACK:102400000,102400000")
15 #define Local
16 #define o(a) cout << #a << "  =  " << a << endl
17 #define pb(a) push_back(a)
18 #define lson l,mid,pos<<1
19 #define rson mid+1,r,pos<<1|1
20 typedef long long ll;
21 using namespace std;
22 
23 void local(){
24 #ifdef Local
25     freopen("date.in","r",stdin);
26     freopen("date.out","w",stdout);
27 #endif
28 }
29 const int maxn = 550;
30 bool debug = false;
31 int n, k;
32 int arr[maxn];
33 priority_queue<int ,vector<int>,greater<int> > q;
34 int main(){
35     int T;
36     scanf("%d", &T);
37 
38     while(T--){
39         scanf("%d%d", &n, &k);
40         int v;
41         while(!q.empty()) q.pop();
42         for(int i = 1; i <= n ; ++i){
43             scanf("%d", &v);
44             q.push(v);
45         }
46         int top;
47         int c = 1;
48         if(debug){ o(q.top());}
49         while(!q.empty() && q.top() <= n){
50             top = q.top();
51             q.pop();
52             if(debug){ o(top);}
53             if(c != top){
54                 q.push(top + k);
55             }else {
56             c++;
57             }
58 
59             //cout << "OK" << endl;
60         }
61         if(!q.empty()){ cout << "Tom" << endl;}
62         else cout << "Jerry" << endl;
63     }
64 }

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