Marriage Match IV
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5225 Accepted Submission(s): 1549
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 2Sample Output
2
1
1题目大意：给你一个图，问给出的源点到给出的汇点中有多少条最短路，这些最短路之间不能有公共边
先从源点跑一遍最短路，再从汇点跑一遍最短路，如果有dis1[u]+dis2[v]+edge[i].val==dis[汇点]则说明这条路在最短路上，然后我们就用这条边来建图跑最大流。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxm=2e5+7;
const int maxn=10010;
const int inf=0x3f3f3f3f;
struct SPFA
{
int from;
int to;
int val;
int next;
}vec[maxm<<2];
struct Node
{
int to;
int capa;
int next;
}edge[maxm<<2];
int cnt;
int head[maxn];
int head1[maxn];
int cnt1;
int dis[maxn];
int dis1[maxn];
int dis2[maxn];
bool vis[maxn];
int source,sink;
int u[maxm];
int v[maxm];
int w[maxm];
int dep[maxn];
void init()
{
memset(head1,-1,sizeof(head1));
memset(head,-1,sizeof(head));
cnt=0;
cnt1=0;
return;
}
void add(int u,int v,int capa)
{
edge[cnt].to=v;
edge[cnt].capa=capa;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].to=u;
edge[cnt].capa=0;
edge[cnt].next=head[v];
head[v]=cnt++;
return;
}
void add1(int u,int v,int w)
{
vec[cnt1].from=u;
vec[cnt1].to=v;
vec[cnt1].val=w;
vec[cnt1].next=head1[u];
head1[u]=cnt1++;
return;
}
void spfa(int node)
{
memset(dis,inf,sizeof(dis));
memset(vis,false,sizeof(vis));
dis[node]=0;
vis[node]=true;
queue<int> que;
que.push(node);
while(!que.empty())
{
int now=que.front();
que.pop();
vis[now]=false;
for(int i=head1[now];~i;i=vec[i].next)
{
int v=vec[i].to;
if(dis[v]>dis[now]+vec[i].val)
{
dis[v]=dis[now]+vec[i].val;
if(!vis[v])
{
vis[v]=true;
que.push(v);
}
}
}
}
return;
}
bool bfs()
{
queue<int> que;
que.push(source);
memset(dep,-1,sizeof(dep));
dep[source]=0;
while(!que.empty())
{
int node=que.front();
que.pop();
for(int i=head[node];~i;i=edge[i].next)
{
int v=edge[i].to;
if(edge[i].capa>0&&dep[v]==-1)
{
dep[v]=dep[node]+1;
if(v==sink) return true;
que.push(v);
}
}
}
return dep[sink]!=-1;
}
int dfs(int node,int minn)
{
if(node==sink||minn==0)
{
return minn;
}
int r=0;
for(int i=head[node];~i;i=edge[i].next)
{
int v=edge[i].to;
if(edge[i].capa>0&&dep[v]==dep[node]+1)
{
int tmp=dfs(v,min(edge[i].capa,minn));
if(tmp>0)
{
edge[i].capa-=tmp;
edge[i^1].capa+=tmp;
r+=tmp;
minn-=tmp;
if(!minn) break;
}
}
}
if(!r) dep[node]=-1;
return r;
}
int dinic()
{
int maxflow=0;
while(bfs())
{
maxflow+=dfs(source,inf);
}
return maxflow;
}
int main()
{
int test;
scanf("%d",&test);
while(test--)
{
init();
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&u[i],&v[i],&w[i]);
add1(u[i],v[i],w[i]);
}
scanf("%d%d",&source,&sink);
spfa(source);
memcpy(dis1,dis,sizeof(dis));
init();
for(int i=0;i<m;i++)
{
add1(v[i],u[i],w[i]);
}
spfa(sink);
memcpy(dis2,dis,sizeof(dis));
for(int i=0;i<m;i++)
{
if(dis1[u[i]]+dis2[v[i]]+w[i]==dis1[sink])
{
add(u[i],v[i],1);
}
}
printf("%d\n",dinic());
}
return 0;
}