• POJ 3204 Ikki's Story I Road Reconstruction


    POJ_3204

        这个题目和ZOJ_2532是一样的题目,具体的思路可以参考我的ZOJ_2532的解题报告:http://www.cnblogs.com/staginner/archive/2012/08/11/2633751.html

    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #define MAXD 510
    #define MAXM 10010
    #define INF 0x3f3f3f3f
    int N, M, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM];
    int S, T, d[MAXD], q[MAXD], work[MAXD], visx[MAXD], visy[MAXD];
    void add(int x, int y, int z)
    {
        v[e] = y, flow[e] = z;
        next[e] = first[x], first[x] = e ++;
    }
    void init()
    {
        int i, x, y, z;
        S = 0, T = N - 1;
        memset(first, -1, sizeof(first[0]) * N), e = 0;
        for(i = 0; i < M; i ++)
        {
            scanf("%d%d%d", &x, &y, &z);
            add(x, y, z), add(y, x, 0);
        }
    }
    int bfs()
    {
        int i, j, rear = 0;
        memset(d, -1, sizeof(d[0]) * N);
        d[S] = 0, q[rear ++] = S;
        for(i = 0; i < rear; i ++)
            for(j = first[q[i]]; j != -1; j = next[j])
                if(flow[j] && d[v[j]] == -1)
                {
                    d[v[j]] = d[q[i]] + 1, q[rear ++] = v[j];
                    if(v[j] == T) return 1;
                }
        return 0;
    }
    int dfs(int cur, int a)
    {
        if(cur == T) return a;
        for(int &i = work[cur]; i != -1; i = next[i])
            if(flow[i] && d[v[i]] == d[cur] + 1)
                if(int t = dfs(v[i], std::min(a, flow[i])))
                {
                    flow[i] -= t, flow[i ^ 1] += t;
                    return t;
                }
        return 0;
    }
    void dinic()
    {
        while(bfs())
        {
            memcpy(work, first, sizeof(first[0]) * N);
            while(dfs(S, INF));
        }
    }
    void DFS(int cur, int *vis, int k)
    {
        int i;
        vis[cur] = 1;
        for(i = first[cur]; i != -1; i = next[i])
            if(!vis[v[i]] && flow[i ^ k] != 0)
                DFS(v[i], vis, k);
    }
    void solve()
    {
        int i, cnt = 0;
        dinic();
        memset(visx, 0, sizeof(visx[0]) * N), memset(visy, 0, sizeof(visy[0]) * N);
        DFS(S, visx, 0), DFS(T, visy, 1);
        for(i = 0; i < e; i += 2)
            if(flow[i] == 0 && visx[v[i ^ 1]] && visy[v[i]])
                ++ cnt;
        printf("%d\n", cnt);
    }
    int main()
    {
        while(scanf("%d%d", &N, &M) == 2)
        {
            init();
            solve();
        }
        return 0;
    }
  • 相关阅读:
    Statement
    打印页数设定
    点选TOP后并不是直接跳到页顶的,而是滚动上去
    文本框不允许输入特殊字符,只能是数字、字母、-和_,不允许输入空格键
    不间断滚动
    无限级别的菜单(侧拉菜单)
    筛法求素数
    1212
    触发器引发的entityframework异常
    using crystalreport generate PDF2
  • 原文地址:https://www.cnblogs.com/staginner/p/2644957.html
Copyright © 2020-2023  润新知