Given a weighted directed graph, we define the shortest path as the path who has the smallest length among all the path connecting the source vertex to the target vertex. And if two path is said to be non-overlapping, it means that the two path has no common edge. So, given a weighted directed graph, a source vertex and a target vertex, we are interested in how many non-overlapping shortest path could we find out at most.
Input
Input consists of multiple test cases. The first line of each test case, there is an integer number N (1<=N<=100), which is the number of the vertices. Then follows an N * N matrix, represents the directed graph. Each element of the matrix is either non-negative integer, denotes the length of the edge, or -1, which means there is no edge. At the last, the test case ends with two integer numbers S and T (0<=S, T<=N-1), that is, the starting and ending points. Process to the end of the file.
Output
For each test case, output one line, the number of the the non-overlapping shortest path that we can find at most, or "inf" (without quote), if the starting point meets with the ending.
Sample Input
4 0 1 1 -1 -1 0 1 1 -1 -1 0 1 -1 -1 -1 0 0 3 5 0 1 1 -1 -1 -1 0 1 1 -1 -1 -1 0 1 -1 -1 -1 -1 0 1 -1 -1 -1 -1 0 0 4
Sample Output
2 1
题意：给定源点和汇点的情况下边不相交最短路的条数
题解：先用floyed算法将图中的点全部两点之间的距离全部算一遍，求出最短路，然后枚举原来图中的边，如果在最短路上，则连一条容量为1的边，算出最大流即可。
//zoj 2760 #include <stdio.h> #include <algorithm> #include <queue> #include <string.h> #include <math.h> #include <iostream> #include <math.h> using namespace std; const int N = 205; const int INF = 999999999; struct Edge{ int v,next; int w; }edge[2*N*N]; int head[N]; int level[N]; int graph[N][N],dis[N][N]; int tot; void init(){ memset(head,-1,sizeof(head)); tot=0; } void addEdge(int u,int v,int w,int &k){ edge[k].v = v,edge[k].w=w,edge[k].next=head[u],head[u]=k++; edge[k].v = u,edge[k].w=0,edge[k].next=head[v],head[v]=k++; } int BFS(int src,int des){ queue<int >q; memset(level,0,sizeof(level)); level[src]=1; q.push(src); while(!q.empty()){ int u = q.front(); q.pop(); if(u==des) return 1; for(int k = head[u];k!=-1;k=edge[k].next){ int v = edge[k].v; int w = edge[k].w; if(level[v]==0&&w!=0){ level[v]=level[u]+1; q.push(v); } } } return -1; } int dfs(int u,int des,int increaseRoad){ if(u==des) return increaseRoad; int ret=0; for(int k=head[u];k!=-1;k=edge[k].next){ int v = edge[k].v; int w = edge[k].w; if(level[v]==level[u]+1&&w!=0){ int MIN = min(increaseRoad-ret,w); w = dfs(v,des,MIN); if(w>0){ edge[k].w -=w; edge[k^1].w+=w; ret+=w; if(ret==increaseRoad) return ret; }else level[v] = -1; } } return ret; } int Dinic(int src,int des){ int ans = 0; while(BFS(src,des)!=-1) ans+=dfs(src,des,INF); return ans; } void floyd(int n){ for(int k=0;k<n;k++){ for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ dis[i][j] = min(dis[i][k]+dis[k][j],dis[i][j]); } } } } void build(int s,int t,int N){ int i,j; for(i=0;i<N;i++){ if(dis[s][i]==INF)continue; for(j=0;j<N;j++){ if(i==j||dis[j][t]==INF||dis[i][j]==INF)continue; if(dis[s][t]==dis[s][i]+graph[i][j]+dis[j][t]){ addEdge(i,j,1,tot); } } } } int main(){ int n; while(scanf("%d",&n)!=EOF){ init(); for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ int w; scanf("%d",&graph[i][j]); if(graph[i][j]==-1) graph[i][j]=INF; if(i==j) graph[i][j] = 0; dis[i][j] = graph[i][j]; } } int src,des; scanf("%d%d",&src,&des); if(src==des) { printf("inf ");continue; } floyd(n); build(src,des,n); int ans = Dinic(src,des); printf("%d ",ans); } }