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Luogu P2598 [ZJOI2009]狼和羊的故事

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题目描述

https://www.luogu.com.cn/problem/P2598

简要题意：给定一张 $n\times m$ 的四连通网格图，每个点属于狼、羊和空地三者之一，空地可以连接羊和狼的领地，求最少需要割掉多少条边，使得狼和羊分开

$n,m\le 100$

Solution

注意到要割掉的边一定是 羊 空地 狼

建图：

$s$ 连 羊，容量为 $\infty$

羊 连 空地/羊，容量为 $1$

羊/空地 连 狼，容量为 $1$

狼 连 $t$，容量为 $\infty$

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#include <iostream>
#include <cstdio>
#include <queue>
#define maxn 10010
#define maxm 110
#define INF 1000000000
using namespace std;

int n, m, a[maxm][maxm];

int dx[] = { 0, 0, -1, 1 };
int dy[] = { 1, -1, 0, 0 };

inline int id(int x, int y) { return (x - 1) * m + y; } 

struct Edge {
    int to, next, w;
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
    e[c1].to = v; e[c1].w = w;
    e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w) {
    add_edge(u, v, w); add_edge(v, u, 0);
}

int dep[maxn], s, t; 
bool bfs() {
    fill(dep, dep + maxn, 0); dep[s] = 1;
    queue<int> Q; Q.push(s);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to, w = e[i].w;
            if (!dep[v] && w > 0) {
                dep[v] = dep[u] + 1;
                Q.push(v); if (v == t) return 1;
            }
        }
    }
    return 0; 
} 

int dfs(int u, int _w) {
    if (!_w || u == t) return _w;
    int s = 0;
    for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to, w = e[i].w;
        if (w > 0 && dep[v] == dep[u] + 1) {
            int di = dfs(v, min(_w - s, w));
            e[i].w -= di; e[i ^ 1].w += di;
            s += di; if (s == _w) break;
        }
    }
    if (s < _w) dep[u] = 0;
    return s; 
}

int mf;
int dinic() {
    while (bfs()) mf += dfs(s, INF);
    return mf;
} 

int main() { fill(head, head + maxn, -1); 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> m; s = 0; t = n * m + 1; 
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j) cin >> a[i][j];
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j) {
            if (a[i][j] == 1) Add_edge(s, id(i, j), INF);
            else if (a[i][j] == 2) { Add_edge(id(i, j), t, INF); continue; } 
            for (int k = 0; k < 4; ++k) {
                int x = i + dx[k], y = j + dy[k];
                if (x < 1 || x > n || y < 1 || y > m || a[x][y] == 1) continue;
                Add_edge(id(i, j), id(x, y), 1);
            } 
        }
    cout << dinic() << "\n"; 
    return 0;
}