hdu 3416 Marriage Match IV【SPFA+最大流Dinic】

Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3417    Accepted Submission(s): 1019


Problem Description

Do not sincere non-interference。

Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can get to city B and make a data with a girl he likes. But there are two problems with it, one is
starvae must get to B within least time, it's said that he must take a shortest path. Other is no road can be taken more than once. While the city starvae passed away can been taken more than once.

So, under a good RP, starvae may have many chances to get to city B. But he don't know how many chances at most he can make a data with the girl he likes . Could you help starvae?

Input

The first line is an integer T indicating the case number.(1<=T<=65)

For each case,there are two integer n and m in the first line ( 2<=n<=1000, 0<=m<=100000 ) ,n is the number of the city and m is the number of the roads.

Then follows m line ,each line have three integers a,b,c,(1<=a,b<=n,0<c<=1000)it means there is a road from a to b and it's distance is c, while there may have no road from b to a. There may have a road from a to a,but you can ignore it. If there are two roads
from a to b, they are different.

At last is a line with two integer A and B(1<=A,B<=N,A!=B), means the number of city A and city B.

There may be some blank line between each case.

Output

Output a line with a integer, means the chances starvae can get at most.

Sample Input

3

7 8

1 2 1

1 3 1

2 4 1

3 4 1

4 5 1

4 6 1

5 7 1

6 7 1

1 7

 

6 7

1 2 1

2 3 1

1 3 3

3 4 1

3 5 1

4 6 1

5 6 1

1 6

 

2 2

1 2 1

1 2 2

1 2

 

 

Sample Output

2

1

1

 

 

Author

starvae@HDU

 

题目大意:

图中有n个点，m条有向边，然后给你起点和终点，问你从起点到终点的最短路数为多少条，需要保证每条路没有重复的边。

思路：

1、对于这样一个理论不难想到：

①我们求一遍单源最短路、

如果有：dis【from】+当前边权值==dis【to】，那么当前边就有可能是整条最短路径中的某一条边，那么当前边就有可能成为某一种走法的一条路径。

②那么我们就先求一遍单源最短路，然后枚举每一条边，如果这条边符合上述条件，那么将这条边加入网络，并设定其流量为1.

2、对于刚刚处理完的一个网络，能够保证从u到v随便怎么走都是一条最短路，那么我们直接跑最大流即可，此时最大流得到的值，就是最大最短路径数。

Ac代码：

#include<stdio.h>
#include<queue>
#include<iostream>
#include<string.h>
#include<vector>
using namespace std;
#define INF 0x3f3f3f3f
struct node2
{
int v;int w;
}now;
struct node
{
int from;
int to;
int w;
int next;
}e[1515155],ee[151555];
int divv[1500];
int head[1500];
int head2[1500];
int cur[1500];
int dis[1055];
int vis[1055];
int n,m;
int cont,ss,tt,cont2;
void add2(int from,int to,int w)
{
ee[cont2].to=to;
ee[cont2].from=from;
ee[cont2].w=w;
ee[cont2].next=head2[from];
head2[from]=cont2++;
}
void add(int from,int to,int w)
{
e[cont].from=from;
e[cont].to=to;
e[cont].w=w;
e[cont].next=head[from];
head[from]=cont++;
}
void SPFA()
{
for(int i=1;i<=n;i++)dis[i]=0x3f3f3f3f;
dis[ss]=0;
vis[ss]=1;
queue<int >s;
s.push(ss);
while(!s.empty())
{
int u=s.front();
s.pop();vis[u]=0;
for(int i=head[u];i!=-1;i=e[i].next)
{
int v=e[i].to;
int w=e[i].w;
if(dis[v]>dis[u]+w)
{
dis[v]=dis[u]+w;
if(vis[v]==0)
{
s.push(v);
vis[v]=1;
}
}
}
}
}
int makedivv()
{
memset(divv,0,sizeof(divv));
divv[ss]=1;
queue<int >s;
s.push(ss);
while(!s.empty())
{
int u=s.front();s.pop();
if(u==tt)return 1;
for(int i=head2[u];i!=-1;i=ee[i].next)
{
int v=ee[i].to;
int w=ee[i].w;
if(w&&divv[v]==0)
{
divv[v]=divv[u]+1;
s.push(v);
}
}
}
return 0;
}
int Dfs(int u,int maxflow,int tt)
{
if(u==tt)return maxflow;
int ret=0;
for(int &i=cur[u];i!=-1;i=ee[i].next)
{
int v=ee[i].to;
int w=ee[i].w;
if(w&&divv[v]==divv[u]+1)
{
int f=Dfs(v,min(maxflow-ret,w),tt);
ee[i].w-=f;
ee[i^1].w+=f;
ret+=f;
if(ret==maxflow)return ret;
}
}
return ret;
}
void Dinic()
{
int ans=0;
while(makedivv()==1)
{
memcpy(cur,head2,sizeof(head2));
ans+=Dfs(ss,INF,tt);
}
printf("%d\n",ans);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
cont=0;
cont2=0;
memset(head2,-1,sizeof(head2));
memset(head,-1,sizeof(head));
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
int x,y,w;
scanf("%d%d%d",&x,&y,&w);
add(x,y,w);
}
scanf("%d%d",&ss,&tt);
SPFA();
for(int i=0;i<cont;i++)
{
int u=e[i].from;
int v=e[i].to;
int w=e[i].w;
if(dis[u]+w==dis[v]&&w)
{
add2(u,v,1);
add2(v,u,0);
}
}
Dinic();
}
}