Silver Cow Party（最短路，好題）

Silver Cow Party

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 3268

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively:  N,  M, and  X 
Lines 2.. M+1: Line  i+1 describes road  i with three space-separated integers:  A_i,  B_i, and  T_i. The described road runs from farm  A_i to farm  B_i, requiring  T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

 

求兩次最短路，第一次求x到其他各點的最短路，第二次求各點到x的最短路。前者易於解決，直接應用spfa或其餘最短路算法便可，後者要先將鄰接矩陣轉置再執行最短路算法。

爲何進行矩陣轉置？好比u（u ！= x）到x的最短路爲<u,v1>,<v1,v2>,<v2,v3>,...,<vi, x>，這條路徑在轉置鄰接矩陣後變成<x,vi>,...,<v3,v2>,<v2, v1>,<v1,u>.因而乎，在轉置鄰接矩陣後，執行最短路算法求出x到u的最短路<x,vi>,...,<v3,v2>,<v2, v1>,<v1,u>便可獲得轉置前u到x的最短路。

 

 1     #include <iostream>
 2     #include <deque>
 3     #include <cstdio>
 4     #include <cstring>
 5     #include <algorithm>
 6 
 7     using namespace std;
 8 
 9     const int MAXV = 1002;
10     const int inf = 0x3f3f3f3f;
11     int t[MAXV][MAXV], d1[MAXV], d2[MAXV];
12     int que[MAXV<<1];
13     bool in[MAXV];
14     int n, m, x;
15 
16     void spfa(int * d)
17     {
18         memset(in, false, sizeof(in));
19         memset(d + 1, inf, sizeof(int) * n);//memset(d, inf, sizeof(d)) if wrong
20         d[x] = 0;
21         int tail = -1;
22         que[++tail] = x;
23         in[x] = true;
24         while(tail != -1){
25             int cur = que[tail];
26             tail--;
27             in[cur] = false;
28             for(int i = 1; i <= n; i++){
29                 if(d[cur] + t[cur][i] < d[i]){
30                     d[i] = d[cur] + t[cur][i];
31                     if(in[i] == false){
32                         que[++tail] = i;
33                         in[i] = true;
34                     }
35                 }
36             }
37         }
38     }
39 
40     void tran()
41     {
42         int i, j;
43         for(i = 1; i <= n; i++){
44             for(j = 1; j <= i; j++){
45                 swap(t[i][j], t[j][i]);
46             }
47         }
48     }
49 
50     int main()
51     {
52         while(scanf("%d %d %d", &n, &m, &x) != EOF){
53             memset(t, inf, sizeof(t));
54             while(m--){
55                 int a, b, c;
56                 scanf("%d %d %d", &a, &b, &c);
57                 t[a][b] = c;
58             }
59             spfa(d1);
60             tran();
61             spfa(d2);
62             int ans = -1;
63             for(int i = 1; i <= n; i++){
64                 if(d1[i] != inf && d2[i] != inf)
65                     ans = max(ans, d1[i] + d2[i]);
66             }
67             printf("%d\n", ans);
68         }
69         return 0;
70     }