题意 ： 给你两个城市让你求最短距离，如果两个城市位于同一强连通分量中那距离为0.
思路 ：强连通分量缩点之后，求最短路。以前写过，总感觉记忆不深，这次自己敲完再写了一遍。
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <vector> 5 #include <stack> 6 #include <queue> 7 #include <algorithm> 8 #define maxn 505 9 using namespace std ; 10 11 struct node 12 { 13 int u,v,w,next ; 14 } edge[maxn*maxn]; 15 int p[maxn][maxn]; 16 int dfn[maxn],low[maxn],head[maxn] ,vis[maxn],dis[maxn],belong[maxn]; 17 int timee,cnt ,cntt,n,m; 18 stack<int>stk ; 19 20 void Init() 21 { 22 timee = 0 ; 23 cnt = cntt = 0 ; 24 memset(dfn,0,sizeof(dfn)) ; 25 memset(low,0,sizeof(low)) ; 26 memset(dis,0,sizeof(dis)) ; 27 memset(head,-1,sizeof(head)) ; 28 memset(vis,0,sizeof(vis)) ; 29 memset(p,0x3f,sizeof(p)) ; 30 } 31 void addedge(int u,int v,int w) 32 { 33 edge[cnt].u = u ; 34 edge[cnt].v = v ; 35 edge[cnt].w = w ; 36 edge[cnt].next = head[u] ; 37 head[u] = cnt++ ; 38 } 39 void tarjan(int u) 40 { 41 //cout<<u<<endl; 42 int v ; 43 vis[u] = 1 ; 44 dfn[u] = low[u] = ++ timee ; 45 stk.push(u) ; 46 for(int i = head[u] ; i != -1 ; i = edge[i].next) 47 { 48 v = edge[i].v ; 49 if(!dfn[v]) 50 { 51 //printf("v = %d ",v) ; 52 tarjan(v) ; 53 low[u] = min(low[v],low[u]) ; 54 } 55 else if(vis[v]) 56 low[u] = min(low[u],dfn[v]) ; 57 } 58 if(low[u] == dfn[u]) 59 { 60 cntt ++ ; 61 do 62 { 63 v = stk.top() ; 64 stk.pop() ; 65 vis[v] = 0 ; 66 belong [v] = cntt ; 67 // puts("1") ; 68 } 69 while(v != u) ; 70 } 71 } 72 void SPFA(int u,int v) 73 { 74 queue<int>Q ; 75 for(int i = 1 ; i <= cntt ; i++) 76 { 77 dis[i] = 999999999 ; 78 vis[i] = 0 ; 79 } 80 dis[u] = 0; 81 vis[u] = 1; 82 Q.push(u) ; 83 while(!Q.empty()) 84 { 85 int s = Q.front() ; 86 Q.pop() ; 87 vis[s] = 0 ; 88 for(int i = 1 ; i <= n ; i ++ ) 89 { 90 if(p[s][i] != 999999999) 91 { 92 if(dis[i] > dis[s] + p[s][i]) 93 { 94 dis[i] = dis[s] + p[s][i] ; 95 if(!vis[i]) 96 { 97 Q.push(i) ; 98 vis[i] = 1 ; 99 } 100 } 101 } 102 } 103 } 104 if(dis[v] != 999999999) 105 printf("%d ",dis[v]) ; 106 else printf("Nao e possivel entregar a carta ") ; 107 } 108 void rebuild() 109 { 110 for(int i = 1 ; i <= n ; i++) 111 { 112 for(int j = head[i] ; j != -1 ; j = edge[j].next) 113 { 114 // int s = edge[i].v ; 115 int v = belong[edge[j].v] ; 116 int u = belong[edge[j].u] ; 117 if(u != v) 118 p[u][v] = min(p[u][v],edge[j].w) ; 119 } 120 } 121 for(int i = 1 ; i <= cntt ; i++) 122 p[i][i] = 0 ; 123 } 124 int main() 125 { 126 int x,y,h,k; 127 while(~scanf("%d %d",&n,&m)) 128 { 129 if(n == 0 && m == 0) break ; 130 Init() ; 131 for(int i = 0 ; i < m ; i++) 132 { 133 scanf("%d %d %d",&x,&y,&h) ; 134 addedge(x,y,h) ; 135 } 136 for(int i = 1 ; i <= n ; i++) 137 { 138 if(!dfn[i]) 139 tarjan(i) ; 140 } 141 // cout<<"s"<<endl; 142 rebuild() ; 143 // cout<<"s"<<endl; 144 scanf("%d",&k) ; 145 for(int i = 0 ; i < k ; i++) 146 { 147 //cout<<i<<endl; 148 scanf("%d %d",&x,&y) ; 149 SPFA(belong[x],belong[y]) ; 150 } 151 printf(" ") ; 152 } 153 return 0 ; 154 }