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Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6230 Accepted Submission(s): 1804

題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=3416

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

題意：

一個男生要去到一個城市找女生，他們兩個的城市間有多條路徑，現在他肯定想走最短路啦~

問他能走多少條最短路到女生的城市，並且要求這些最短路路徑不能重復。

題解:

最近事情有點多，很久沒更新博客了...

這題一開始我想的是費用流，但是T了。

正解是最大流，但是是經過“改造後”的最大流。這個最大流網絡應該是由容量為1的，在原圖最短路徑上的邊構成，最後我們跑個Dinic就行了。

要找最短路徑上面的邊，我們從起點spfa一次，再從終點spfa一次就行了。註意跑spfa建邊的時候不要建雙向邊，我一開始就是這裏WA了...

具體做法見代碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
#define INF 1<<30
using namespace std;
typedef long long ll;
const int N =1005, M = 1e5+5;
int n,m,s,t,T,mx;
int d1[N],d2[N],d[N],vis[N],head[N];
pair <pair<int,int>,int> E[M];
struct Edge{
    int v,next,c,w;
}e[M<<1];
int tot;
void adde(int u,int v,int c,int w){
    e[tot].v=v;e[tot].c=c;e[tot].next=head[u];e[tot].w=w;head[u]=tot++;
}
void spfa(int S){
    queue <int> q;q.push(S);
    memset(vis,0,sizeof(vis));vis[S]=1;
    for(int i=1;i<=n;i++) d2[i]=INF;d2[S]=0;
    while(!q.empty()){
        int u=q.front();q.pop();vis[u]=0;
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].v;
            if(d2[v]>d2[u]+e[i].w){
                d2[v]=d2[u]+e[i].w;
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}
int bfs(){
    memset(d,0,sizeof(d));d[s]=1;
    queue <int > q;q.push(s);
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].v;
            if(e[i].c>0 && !d[v]){
                d[v]=d[u]+1;
                q.push(v);
            }
        }
    }
    return d[t]!=0;
}
int dfs(int u,int a){
    if(u==t || a==0) return a;
    int flow=0,f;
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].v;
        if(d[v]!=d[u]+1) continue ;
        f=dfs(v,min(a,e[i].c));
        if(f>0){
            e[i].c-=f;
            e[i^1].c+=f;
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    if(!flow) d[u]=-1;
    return flow;
}
int Dinic(){
    int flow=0;
    while(bfs()) flow+=dfs(s,INF);
    return flow;
}
int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        tot=0;memset(head,-1,sizeof(head));
        for(int i=1;i<=m;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            E[i]=make_pair(make_pair(u,v),w);
            adde(u,v,1,w);
        }
        scanf("%d%d",&s,&t);
        spfa(s);
        for(int i=1;i<=n;i++) d1[i]=d2[i];
        tot=0;memset(head,-1,sizeof(head));
        for(int i=1;i<=m;i++){
            int u=E[i].first.first,v=E[i].first.second,w=E[i].second;
            adde(v,u,1,w);
        }
        spfa(t);
        mx=d1[t];
        tot=0;memset(head,-1,sizeof(head));
        for(int i=1;i<=m;i++){
            int u =E[i].first.first,v=E[i].first.second,w=E[i].second;
            if(d1[u]+d2[v]+w==mx){
                adde(u,v,1,INF);
                adde(v,u,0,INF);
            }
        }
        printf("%d\n",Dinic());
    }
    return 0;
}

HDU3605：Marriage Match IV