
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.
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?
题意就是求在最短路的基础上有几条路可以到达,但是每条路之间边彼此不能重合。。。
然后就是先求出最短路来,然后把所以 lowcost[v]==lowcost[u]+cost[u][v] 的边留下,然后再求最大流就好了。。。。。。
代码如下:
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h> using namespace std; const int MaxN=;
const int MaxM=;
const int INF=10e8; namespace first
{ struct Edge
{
int to,next,cost;
}; Edge E[MaxM];
int head[MaxN],Ecou;
int vis[MaxN]; void init(int N)
{
Ecou=; for(int i=;i<=N;++i)
{
head[i]=-;
vis[i]=;
}
} void addEdge(int u,int v,int c)
{
E[Ecou].to=v;
E[Ecou].cost=c;
E[Ecou].next=head[u];
head[u]=Ecou++;
} void SPFA(int lowcost[],int N,int start)
{
queue <int> que;
int u,v,c; for(int i=;i<=N;++i)
lowcost[i]=INF;
lowcost[start]=; que.push(start);
vis[start]=; while(!que.empty())
{
u=que.front();
que.pop(); vis[u]=; for(int i=head[u];i!=-;i=E[i].next)
{
v=E[i].to;
c=E[i].cost; if(lowcost[v]>lowcost[u]+c)
{
lowcost[v]=lowcost[u]+c; if(!vis[v])
{
que.push(v);
vis[v]=;
}
}
}
}
} } namespace second
{ struct Edge
{
int to,next,cap,flow;
}; Edge E[MaxM];
int Ecou,head[MaxN];
int gap[MaxN],dis[MaxN],pre[MaxN],cur[MaxN];
int S,T; void init(int N,int _S,int _T)
{
S=_S;
T=_T;
Ecou=; for(int i=;i<=N;++i)
{
head[i]=-;
gap[i]=dis[i]=;
}
} void addEdge(int u,int v,int c,int rc=)
{
E[Ecou].to=v;
E[Ecou].cap=c;
E[Ecou].flow=;
E[Ecou].next=head[u];
head[u]=Ecou++; E[Ecou].to=u;
E[Ecou].cap=rc;
E[Ecou].flow=;
E[Ecou].next=head[v];
head[v]=Ecou++;
} void update(int remm)
{
int u=T; while(u!=S)
{
E[pre[u]].flow+=remm;
E[pre[u]^].flow-=remm;
u=E[pre[u]^].to;
}
} int SAP(int N)
{
for(int i=;i<=N;++i)
cur[i]=head[i]; int u,v,ret=,remm=INF,mindis; u=S;
pre[S]=-;
gap[]=N; while(dis[S]<N)
{
loop:
for(int i=cur[u];i!=-;i=E[i].next)
{
v=E[i].to; if(E[i].cap-E[i].flow && dis[u]==dis[v]+)
{
pre[v]=i;
cur[u]=i;
u=v; if(u==T)
{
for(int i=pre[u];i!=-;i=pre[E[i^].to])
remm=min(remm,E[i].cap-E[i].flow); ret+=remm;
update(remm);
u=S;
remm=INF;
} goto loop;
}
} mindis=N-;
for(int i=head[u];i!=-;i=E[i].next)
if(E[i].cap-E[i].flow && mindis>dis[E[i].to])
{
cur[u]=i;
mindis=dis[E[i].to];
} if(--gap[dis[u]]==)
break; dis[u]=mindis+; ++gap[dis[u]]; if(u!=S)
u=E[pre[u]^].to;
} return ret;
} } int lowcost[MaxN]; int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout); int T;
int N,M;
int A,B;
int a,b,c; scanf("%d",&T); while(T--)
{
scanf("%d %d",&N,&M); first::init(N); {
using namespace first; while(M--)
{
scanf("%d %d %d",&a,&b,&c); addEdge(a,b,c);
} scanf("%d %d",&A,&B); SPFA(lowcost,N,A); second::init(N,A,B); for(int u=;u<=N;++u)
for(int i=head[u];i!=-;i=E[i].next)
if(lowcost[E[i].to]==lowcost[u]+E[i].cost)
second::addEdge(u,E[i].to,);
} {
using namespace second; printf("%d\n",SAP(N));
}
} return ;
}