• BeiJing2006 狼抓兔子


    HYSBZ_1001

        建好图之后求最小割即可。

    #include<stdio.h>
    #include<string.h>
    #define MAXD 1000100
    #define MAXM 6000600
    #define INF 1000000000
    int N, M, T, e, first[MAXD], next[MAXM], u[MAXM], v[MAXM], flow[MAXM];
    int q[MAXD], d[MAXD], work[MAXD], s[MAXD];
    void add(int a, int b, int f)
    {
    u[e] = a;
    v[e] = b;
    flow[e] = f;
    next[e] = first[a];
    first[a] = e;
    e ++;
    }
    void init()
    {
    int i, j, a, b, f;
    T = N * M - 1;
    e = 0;
    for(i = 0; i <= T; i ++)
    first[i] = -1;
    for(i = 0; i < N; i ++)
    for(j = 1; j < M; j ++)
    {
    scanf("%d", &f);
    a = i * M + j - 1;
    b = i * M + j;
    add(a, b, f);
    add(b, a, f);
    }
    for(i = 1; i < N; i ++)
    for(j = 0; j < M; j ++)
    {
    scanf("%d", &f);
    a = (i - 1) * M + j;
    b = i * M + j;
    add(a, b, f);
    add(b, a, f);
    }
    for(i = 1; i < N; i ++)
    for(j = 1; j < M; j ++)
    {
    scanf("%d", &f);
    a = (i - 1) * M + j - 1;
    b = i * M + j;
    add(a, b, f);
    add(b, a, f);
    }
    }
    int bfs()
    {
    int i, j, rear;
    for(i = 1; i <= T; i ++)
    d[i] = -1;
    d[0] = 0;
    rear = 0;
    q[rear ++] = 0;
    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;
    if(v[j] == T)
    return 1;
    q[rear ++] = v[j];
    }
    return 0;
    }
    int dinic()
    {
    int i, j, cur, r, res = 0;
    while(bfs())
    {
    r = 0;
    cur = 0;
    for(i = 0; i <= T; i ++)
    work[i] = first[i];
    for(;;)
    {
    if(cur == T)
    {
    int a = INF, minr = r;
    for(i = 0; i < r; i ++)
    if(flow[s[i]] < a)
    {
    a = flow[s[i]];
    minr = i;
    }
    for(i = 0; i < r; i ++)
    {
    flow[s[i]] -= a;
    flow[s[i] ^ 1] += a;
    }
    r = minr;
    cur = u[s[r]];
    res += a;
    }
    for(i = work[cur]; i != -1; i = next[i])
    if(flow[i] && d[v[i]] == d[cur] + 1)
    break;
    work[cur] = i;
    if(i != -1)
    {
    s[r ++] = i;
    cur = v[i];
    }
    else
    {
    d[cur] = -1;
    if(r == 0)
    break;
    r --;
    cur = u[s[r]];
    }
    }
    }
    return res;
    }
    int main()
    {
    while(scanf("%d%d", &N, &M) == 2)
    {
    init();
    int res = dinic();
    printf("%d\n", res);
    }
    return 0;
    }


  • 相关阅读:
    百度网盘破解
    openstack2 kvm
    Openstack1 云计算与虚拟化概念
    Rsync + Sersync 实现数据增量同步
    Ansible 详解2-Playbook使用
    Ansible 详解
    Python mysql sql基本操作
    COBBLER无人值守安装
    ELK 环境搭建4-Kafka + zookeeper
    此坑待填 离散化思想和凸包 UVA
  • 原文地址:https://www.cnblogs.com/staginner/p/2225653.html
Copyright © 2020-2023  润新知