【BZOJ1001】[BeiJing2006]狼抓兔子

挺简单一个题,最小割模板

我的感觉就是可能建图的时候会比较麻烦吧,毕竟三个方向。

#include <cctype>
#include <climits>
#include <cstdio>
#include <cstring>
#include <iostream>


#define debug(x) std::cout << #x << " = " << x << std::endl;

#define __OPTIMIZED__INPUT__
int nextInt()
{
    int num = 0;
    char c;
    bool flag = false;
    while ((c = std::getchar()) == ' ' || c == '\r' || c == '\n' || c == '\t');
    if (c == '-')
        flag = true;
    else
        num = c - 48;
    while (std::isdigit(c = std::getchar()))
        num = num * 10 + c - 48;
    return (flag ? - 1 : 1) * num;
}

struct
{
    int to;
    int nex;
    int v;
} e[600001];
int head[600001];
int h[600001], q[600001];
int ans, n, m;

void Insert(const int u, const int v, const int w)
{
    static int tot = 0;
    tot++;
    e[tot].to = v;
    e[tot].v = w;
    e[tot].nex = head[u];
    head[u] = tot;
}

bool bfs()
{
    int now, i;
    std::memset(h, 0xff, sizeof h);
    int t = 0;
    int w = 1;
    q[t] = 1;
    h[1] = 0;
    while (t < w)
    {
        now = q[t];
        t++;
        i = head[now];
        for (int i = head[now]; i; i = e[i].nex)
            if (e[i].v && h[e[i].to] < 0)
            {
                q[w++] = e[i].to;
                h[e[i].to] = h[now] + 1;
            }
    }
    if (h[n * m] == -1)
        return 0;
    return 1;
}

int dfs(const int x, const int f)
{
    if (x == n * m)
        return f;
    int w, used = 0;
    for (int i = head[x]; i; i = e[i].nex)
        if (e[i].v && h[e[i].to] == h[x] + 1)
        {
            w = dfs(e[i].to, std::min(w, e[i].v));
            e[i].v -= w;
            e[i + 1].v += w;
            used += w;
            if (used == f)
                return f;
        }
    if (!used)
        h[x] = -1;
//  debug(used);
    return used;
}

void Dinic()
{
    while (bfs())
        ans += dfs(1, INT_MAX);
}

int main(int argc, char ** argv)
{
    n = nextInt();
    m = nextInt();
    int x;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j < m; j++)
        {
            x = nextInt();
            Insert(m * (i - 1) + j, m * (i - 1) + j + 1, x);
            Insert(m * (i - 1) + j + 1, m * (i - 1) + j, x);
        }
    for (int i = 1; i < n; i++)
        for (int j = 1; j <= m; j++)
        {
            x = nextInt();
            Insert(m * (i - 1) + j, m * i + j, x);
            Insert(m * i + j, m * (i - 1) + j, x);
        }
    for (int i = 1; i < n; i++)
        for (int j = 1; j < m; j++)
        {
            x = nextInt();
            Insert(m * (i - 1) + j, m * i + j + 1, x);
            Insert(m * i + j + 1, m * (i - 1) + j, x);
        }
    Dinic();
    std::cout << ans/* << std::endl*/;
#ifdef __NOTEPADPP
    std::cin.get();
    std::cin.get();
#endif
    return 0;
}
posted @ 2017-03-18 10:45  Edward_Tsui  阅读(115)  评论(0编辑  收藏  举报