题意：给你一个有向权图，问你从S到E有几条最短路，每条边直走一次的情况下；
解题思路：每条边直走一次，最大流边权为1，因为要算几条最短路，那么能得到最短路的路径标记下，然后跑最大流
代码：
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxm=100500;
const int maxn=1050;
struct Edge
{
int next;int to;int w;int fa;
}edge[maxm],edge2[maxm];
struct node
{
int num;
int dist;
node(int _num=0,int _dist=0):num(_num),dist(_dist){}
friend bool operator<(node a,node b)
{
return a.dist>b.dist;
}
};
int head[maxn],head2[maxn],cnt,cnt2;
int dist[maxn],n,m,Start,End;
int depth[maxn];
int x[maxm],y[maxm],w[maxm];
void add(int u,int v,int w)
{
edge[cnt].next=head[u];edge[cnt].to=v;edge[cnt].w=w;head[u]=cnt++;
}
void add2(int u,int v,int w)
{
edge2[cnt2].next=head2[u];edge2[cnt2].to=v;edge2[cnt2].w=w;edge2[cnt2].fa=u;head2[u]=cnt2++;
edge2[cnt2].next=head2[v];edge2[cnt2].to=u;edge2[cnt2].w=0;edge2[cnt2].fa=v;head2[v]=cnt2++;
}
void dij(int x)
{
memset(dist,inf,sizeof(dist));
priority_queue<node>q;q.push(node(x,0));dist[x]=0;
while(!q.empty())
{
node u=q.top();q.pop();int now=u.num;
for(int i=head[now];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dist[v]>dist[now]+edge[i].w)
{
dist[v]=dist[now]+edge[i].w;q.push(node(v,dist[v]));
}
}
}
}
bool bfs()//分层;
{
memset(depth,0,sizeof(depth));
queue<int>q;
q.push(Start);
depth[Start]=1;
while(!q.empty())
{
int temp=q.front();
q.pop();
for(int i=head2[temp];i!=-1;i=edge2[i].next)
{
int v=edge2[i].to;
if(depth[v]||edge2[i].w<=0)
continue;
depth[v]=depth[temp]+1;
q.push(v);
}
}
return depth[End];//若为0表示没法到达也就是没有路径了；
}
int dfs(int u,int maxflow)
{
if(u==End)
return maxflow;
int add=0;
for(int i=head2[u];i!=-1&&add<maxflow;i=edge2[i].next)
{
int v=edge2[i].to;
if(depth[v]!=depth[u]+1)
continue;
if(edge2[i].w==0)
continue;
int tempflow=dfs(v,min(edge2[i].w,maxflow-add));
edge2[i].w-=tempflow;
edge2[i^1].w+=tempflow;
add+=tempflow;
}
return add;
}
int dinic()
{
int ans=0;
while(bfs())
{
ans+=dfs(Start,inf);
}
return ans;
}
int main()
{
int tt;
scanf("%d",&tt);
while(tt--)
{
scanf("%d%d",&n,&m);
memset(head,-1,sizeof(head));memset(head2,-1,sizeof(head2));cnt=cnt2=0;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&x[i],&y[i],&w[i]);add(x[i],y[i],w[i]);
}
scanf("%d%d",&Start,&End);
dij(Start);
for(int i=1;i<=m;i++)
if(dist[y[i]]==dist[x[i]]+w[i])
add2(x[i],y[i],1);
int ans=dinic();
printf("%d
",ans);
}
}