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  • HDU 3416

    Marriage Match IV

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2470    Accepted Submission(s): 742


    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
     

    Source


    题意:
    给出一张无向图,要求图上的不相交得最短路条数。

    思路:
    先求出单源最短路,从终点往起点倒着搜出最短路上的边(能够用双向链表)。然后另这些边的容量为1建新图。跑一遍最大流就可以。


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <queue>
    using namespace std;
    
    #define N 1010
    #define M 101000
    const int INF = 1000000000;
    
    int n, m, sp, tp, cnte, cnte2, head[N], lowdis[N], head2[N], dep[N], tail[N];
    bool vis[N];
    
    class Edge
    {
    public:
        int u, v, w, next, next2;
    };
    Edge edge1[M*2], edge2[M*2];
    
    void addEdge1(int u, int v, int w)
    {
        edge1[cnte].u = u; edge1[cnte].v = v;
        edge1[cnte].w = w; edge1[cnte].next = head[u];
        head[u] = cnte;
        edge1[cnte].next2 = tail[v];
        tail[v] = cnte++;
    }
    void addEdge2(int u, int v, int w)
    {
        edge2[cnte2].u = u; edge2[cnte2].v = v;
        edge2[cnte2].w = w; edge2[cnte2].next = head2[u];
        head2[u] = cnte2++;
    
        edge2[cnte2].v = u; edge2[cnte2].u = v;
        edge2[cnte2].w = 0; edge2[cnte2].next = head2[v];
        head2[v] = cnte2++;
    }
    
    int spfa()
    {
        queue<int> q;
        for(int i = 1; i <= n; i++)
        {
            vis[i] = 0;
            lowdis[i] = INF;
        }
        vis[sp] = 1; lowdis[sp] = 0;
        q.push(sp);
        while(!q.empty())
        {
            int cur = q.front();
            q.pop(); vis[cur] = 0;
            for(int i = head[cur]; i != -1; i = edge1[i].next)
            {
                int v = edge1[i].v;
                if(lowdis[v] > lowdis[cur]+edge1[i].w)
                {
                    lowdis[v] = lowdis[cur]+edge1[i].w;
    
                    if(!vis[v])
                    {
                        vis[v] = 1;
                        q.push(v);
                    }
                }
            }
        }
        return lowdis[tp] != INF;
    }
    
    void dfs(int cur)
    {
        for(int i = tail[cur]; i != -1; i = edge1[i].next2)
        {
            int u = edge1[i].u;
            if(lowdis[u]+edge1[i].w == lowdis[cur])
            {
                addEdge2(u, cur, 1);
                if(!vis[u])
                {
                    vis[u] = 1;
                    dfs(u);
                }
            }
        }
    }
    
    bool bfs()
    {
        memset(vis, 0, sizeof vis);
        memset(dep, -1, sizeof dep);
        queue<int> q;
        q.push(sp);
        vis[sp] = 1;
        dep[sp] = 1;
        while(!q.empty())
        {
            int cur = q.front();
            q.pop();
    
            for(int i = head2[cur]; i != -1; i = edge2[i].next)
            {
                int v = edge2[i].v;
                if(!vis[v] && edge2[i].w > 0)
                {
                    dep[v] = dep[cur]+1;
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    
        return dep[tp] != -1;
    }
    
    int dfs2(int cur, int flow)
    {
        if(cur == tp) return flow;
        int res = 0;
        for(int i = head2[cur]; i != -1 && flow > res; i = edge2[i].next)
        {
            int v = edge2[i].v;
            if(edge2[i].w > 0 && dep[v] == dep[cur]+1)
            {
                int x = min(edge2[i].w, flow-res);
                int f = dfs2(v, x);
                edge2[i].w-=f;
                edge2[i^1].w+=f;
                res += f;
            }
        }
    
        if(!res) dep[cur] = -1;
        return res;
    }
    
    int dinic()
    {
        int res = 0;
        while(bfs())
        {
            int t;
            while(t = dfs2(sp, INF))
            {
                res += t;
            }
        }
        return res;
    }
    
    int main()
    {
    	//freopen("C:\Users\Admin\Desktop\in.txt", "r", stdin);
        int t; scanf("%d", &t);
        while(t--)
        {
            cnte = 0;
            cnte2 = 0;
            memset(head, -1, sizeof head);
            memset(head2, -1, sizeof head2);
            memset(tail, -1, sizeof tail);
            int u, v, w;
            scanf("%d%d", &n, &m);
            for(int i = 0; i < m; i++)
            {
                scanf("%d%d%d", &u, &v, &w);
                addEdge1(u, v, w);
            }
            scanf("%d%d", &sp, &tp);
            int ans;
            if(!spfa()) ans = 0;
            else
            {
                dfs(tp);
                ans = dinic();
            }
            printf("%d
    ", ans);
        }
    	return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/lytwajue/p/6814445.html
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