計蒜客 39280.Travel-二分+最短路dijkstra-二分過程當中保存結果，由於二分完最後的不必定是結果 (The 2019 ACM-ICPC China Shannxi Provincial

時間 2019-11-18

標籤 39280.travel travel 二分 2分短路 dijkstra 過程當中保存結果由於最後必定 acm icpc china shannxi provincial 欄目應用數學简体版

原文原文鏈接

There are $\sim n1∼n. Each planet is connected to other planets through some transmission channels. There are mm transmission channels in the galaxy. Each transmission channel connects two different planets, and each transmission channel has a length.$ node

The residents of the galaxy complete the interplanetary voyage by spaceship. Each spaceship has a level. The spacecraft can be upgraded several times. It can only be upgraded c++

Alice initially has a ide

Input

Each test file contains a single test case. In each test file:oop

The first line contains spa

The second line contains code

Next htm

$\le n\le 10^5, n - 1 \le m \le 10^5,1 \le u,v \le n ,1 \le c,d,e,w \le 10^5)(2≤n≤105,n−1≤m≤105,1≤u,v≤n,1≤c,d,e,w≤105)$ blog

(The graph has no self-loop , no repeated edges , and is connected)ip

Output

Output a line for the minimum cost. Output

樣例輸入

樣例輸出

二分+最短路。

代碼:

  1 //M-二分+最短路dijkstra-二分過程當中保存結果，由於二分完最後的不必定是結果。
  2 #include<bits/stdc++.h>
  3 using namespace std;
  4 typedef long long ll;
  5 const int maxn=1e5+10;
  6 const int inf=0x3f3f3f3f;
  7 #define pii pair<int,int>
  8 #define mp make_pair
  9 #define fi first
 10 #define se second
 11 
 12 namespace IO{
 13     char buf[1<<15],*S,*T;
 14     inline char gc(){
 15         if (S==T){
 16             T=(S=buf)+fread(buf,1,1<<15,stdin);
 17             if (S==T)return EOF;
 18         }
 19         return *S++;
 20     }
 21     inline int read(){
 22         int x; bool f; char c;
 23         for(f=0;(c=gc())<'0'||c>'9';f=c=='-');
 24         for(x=c^'0';(c=gc())>='0'&&c<='9';x=(x<<3)+(x<<1)+(c^'0'));
 25         return f?-x:x;
 26     }
 27     inline long long readll(){
 28         long long x;bool f;char c;
 29         for(f=0;(c=gc())<'0'||c>'9';f=c=='-');
 30         for(x=c^'0';(c=gc())>='0'&&c<='9';x=(x<<3)+(x<<1)+(c^'0'));
 31         return f?-x:x;
 32     }
 33 }
 34 using IO::read;
 35 using IO::readll;
 36 
 37 
 38 int n,m;
 39 int c,d,e;
 40 int head[maxn<<1],cnt;
 41 int dis[maxn],pathmin,edgemin;
 42 bool vis[maxn];
 43 priority_queue<pii,vector<pii>,greater<pii> > q;
 44 
 45 struct Edge{
 46     int u,v,w;
 47 }path[maxn];
 48 
 49 struct node{
 50     int to,next,w;
 51 }edge[maxn<<1];
 52 
 53 void init()
 54 {
 55     memset(head,-1,sizeof head);
 56     memset(edge,0,sizeof edge);
 57     memset(vis,0,sizeof vis);
 58     memset(dis,inf,sizeof dis);
 59     cnt=0;
 60 }
 61 
 62 void add(int u,int v,int w)
 63 {
 64     edge[cnt].to=v;
 65     edge[cnt].w=w;
 66     edge[cnt].next=head[u];
 67     head[u]=cnt++;
 68 }
 69 
 70 void dijkstra(int s)
 71 {
 72     dis[s]=0;
 73     q.push(mp(0,s));
 74     while(!q.empty()){
 75         int u=q.top().se;q.pop();
 76         vis[u]=true;
 77         for(int i=head[u];~i;i=edge[i].next){
 78             int v=edge[i].to;
 79             int w=edge[i].w;
 80             if(dis[v]>dis[u]+w){
 81                 dis[v]=dis[u]+w;
 82                 q.push(mp(dis[v],v));
 83             }
 84         }
 85     }
 86 }
 87 
 88 void solve()
 89 {
 90     pathmin=inf;edgemin=inf;
 91     int l=1,r=1e5;
 92     while(l<=r){
 93         int mid=(l+r)>>1;
 94         init();
 95         for(int i=1;i<=m;i++){
 96             if(path[i].w<=mid){
 97                 add(path[i].u,path[i].v,1);
 98                 add(path[i].v,path[i].u,1);
 99             }
100         }
101         dijkstra(1);
102         if(dis[n]<=mid){
103             r=mid-1;
104             if(edgemin>mid){
105                 edgemin=mid;
106                 pathmin=dis[n];
107             }
108             else if(edgemin==mid){
109                 pathmin=min(pathmin,dis[n]);
110             }
111         }
112         else{
113             l=mid+1;
114         }
115     }
116 }
117 
118 int main()
119 {
120     n=read();m=read();
121 //    scanf("%d%d",&n,&m);
122     c=read();d=read();e=read();
123 //    scanf("%d%d%d",&c,&d,&e);
124     for(int i=1;i<=m;i++){
125         path[i].u=read();path[i].v=read();path[i].w=read();
126 //        scanf("%d%d%d",&path[i].u,&path[i].v,&path[i].w);
127     }
128     solve();//路徑的最大邊權
129     edgemin=edgemin/d+(edgemin%d==0? 0:1);
130     pathmin=pathmin/e+(pathmin%e==0? 0:1); //路徑條數
131     printf("%lld\n",1ll*max(edgemin,pathmin)*c);
132 }
133 */
134 /*
135 
136 5 7
137 1 1 1
138 1 2 3
139 1 3 4
140 1 4 4
141 2 3 2
142 2 4 5
143 3 4 1
144 3 5 3
145 
146 3
147 
148 
149 
150 /*
151 //M-HY學長的代碼orz
152 //二分+最短路spfa
153 #include<bits/stdc++.h>
154 #define  INF  0x3f3f3f3f
155 using namespace std;
156 
157 typedef  long long   ll;
158 const int N = 200010, mod = 1e9+7, M = 300005;
159 
160 ll addCost, addDist, addNum;
161 
162 int n, m, fa[N], head[N], tot2;
163 pair<ll, pair<int ,int > > e1[M];
164 struct Edge{ int to, next; ll w; } e[M];
165 void Init() {
166     tot2=0;
167     memset(head, -1, sizeof(head));
168 }
169 void Addedge(int u, int v, ll w) {
170     e[tot2]={v, head[u], w}; head[u]=tot2++;
171     e[tot2]={u, head[v], w}; head[v]=tot2++;
172 }
173 
174 queue<int > Q;
175 bool inQ[N];
176 ll dis[N];
177 bool spfa(int mid)
178 {
179     memset(inQ, false, sizeof(inQ));
180     memset(dis, INF, sizeof(dis));
181     dis[1] = 0;
182     Q.push(1);
183     inQ[1] = true;
184     ll mostDist = addDist * mid;
185     ll mostNum = addNum * mid;
186     while (!Q.empty())
187     {
188         int u = Q.front();  //cout << u << "   " << dis[u] << "   ---- \n";
189         Q.pop();
190         for (int i = head[u]; i != -1; i = e[i].next)
191         {
192             if (e[i].w > mostDist)
193                 continue;
194             if (dis[e[i].to] > dis[u] + 1)
195             {
196                 dis[e[i].to] = dis[u] + 1;
197                 if (!inQ[e[i].to])
198                 {
199                     Q.push(e[i].to);
200                     inQ[e[i].to] = true;
201                 }
202             }
203         }
204         inQ[u] = false;
205     }
206     return dis[n] != INF && dis[n] <= mostNum;
207 }
208 
209 int main()
210 {
211     scanf("%d %d", &n, &m);
212     scanf("%lld %lld %lld", &addCost, &addDist, &addNum);
213     Init();
214     for (int i = 1; i <= m; ++ i)
215     {
216         scanf("%d %d %lld", &e1[i].second.first, &e1[i].second.second, &e1[i].first);
217         Addedge(e1[i].second.first, e1[i].second.second, e1[i].first);
218     }
219 
220     //cout << 1111 << endl;
221     // 二分
222     int l = 0, r = N, mid, ans = INF;
223     while (l <= r)
224     {
225         mid = (l + r) / 2;
226         if (spfa(mid))
227             ans = mid, r = mid -1;
228         else
229             l = mid + 1;
230         //cout << l << "   " << r << "  " << ans << endl;
231     }
232     if (ans == INF)
233         printf ("-1\n");
234     else
235         printf ("%lld\n", addCost * ans);
236 
237     return 0;
238 }
239 */