HDU 3416 Marriage Match IV

2011-05-17

21:01:42

Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 709 Accepted Submission(s): 188


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题意：求完全不同的最短路数目；题解：先SPFA再用处于最短路上的边建图，容量1； 
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
const int inf=0x3f3f3f3f;
struct rec
{
int u,v,w,link;
}edge[300005];
struct re
{
int v,c;
};
vector<re> E[1005];
bool vis[1005];
int head[1005],pre[1005],cur[1005],dis[1005],gap[1005];
int n,m,num,st,ed;
inline void addedge(int u,int v,int w)
{
edge[num].u=u;
edge[num].v=v;
edge[num].w=w;
edge[num].link=head[u];
head[u]=num++;
edge[num].u=v;
edge[num].v=u;
edge[num].w=0;
edge[num].link=head[v];
head[v]=num++;
}
void SPFA()
{
int i,v,u;
memset(vis,0,sizeof(vis));
for(i=1;i<=n;i++)
dis[i]=inf;
queue<int> Q;
dis[st]=0;
Q.push(st);
while(!Q.empty())
{
u=Q.front();
Q.pop();
vis[u]=0;
for(i=0;i<E[u].size();i++)
{
v=E[u][i].v;
if(dis[v]>dis[u]+E[u][i].c)
{
dis[v]=dis[u]+E[u][i].c;
if(!vis[v])
{
vis[v]=1;
Q.push(v);
}
}
}
}
}
int max_flow()
{
int i,k,u,v,aug=inf,ans=0;
memset(dis,0,sizeof(dis));
memset(gap,0,sizeof(gap));
gap[0]=n;
u=pre[st]=st;
for(i=1;i<=n;i++)
cur[i]=head[i];
while(dis[st]<n)
{
loop:	for(i=cur[u];i;i=cur[u]=edge[i].link)
{
v=edge[i].v;
if(dis[u]==dis[v]+1 && edge[i].w)
{
pre[v]=u;
u=v;
if(aug>edge[i].w)
aug=edge[i].w;
if(v==ed)
{
for(u=pre[v];v!=st;v=u,u=pre[u])
{
edge[cur[u]].w-=aug;
edge[cur[u]^1].w+=aug;
}
ans+=aug;
aug=inf;
}
goto loop;
}
}
k=n;
for(i=head[u];i;i=edge[i].link)
{
v=edge[i].v;
if(k>dis[v] && edge[i].w)
{
cur[u]=i;
k=dis[v];
}
}
if(--gap[dis[u]]==0)
break;
gap[dis[u]=k+1]++;
u=pre[u];
}
return ans;
}
int main()
{
int T,i,j,u,v,c;
re tmp;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
E[i].clear();
num=2;
memset(head,0,sizeof(head));
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&c);
if(u==v)
continue;
tmp.v=v;
tmp.c=c;
E[u].push_back(tmp);
}
scanf("%d%d",&st,&ed);
SPFA();
for(i=1;i<=n;i++)
for(j=0;j<E[i].size();j++)
if(dis[E[i][j].v]==dis[i]+E[i][j].c)
addedge(i,E[i][j].v,1);
/*for(i=2;i<num;i++)
cout<<i<<' '<<edge[i].u<<"-->"<<edge[i].v<<' '<<edge[i].w<<endl;	*/
printf("%d/n",max_flow());
}
return 0;
}