「JSOI2015」圈地

「JSOI2015」圈地

传送门

显然是最小割。

首先对于所有房子，权值 \(> 0\) 的连边 \(s \to i\) ，权值 \(< 0\) 的连边 \(i \to t\) ，然后对于所有的墙，连两条边，连接起墙两边的房子，容量就是修墙的费用，然后直接用权值和 - 最小割就是最大收益。

参考代码：

#include <cstring>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline T min(T a, T b) { return a < b ? a : b; }
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 400 * 400 + 5, __ = 3 * 400 * 400 + 5, INF = 2147483647;

int tot = 1, head[_]; struct Edge { int v, w, nxt; } edge[__ << 1];
inline void Add_edge(int u, int v, int w) { edge[++tot] = (Edge) { v, w, head[u] }, head[u] = tot; }
inline void link(int u, int v, int w) { Add_edge(u, v, w), Add_edge(v, u, 0); }

int n, m, x, sum, s, t, dep[_], cur[_];

inline int bfs() {
    static int hd, tl, Q[_];
    memset(dep, 0, sizeof (int) * (t - s + 1));
    hd = tl = 0, Q[++tl] = s, dep[s] = 1;
    while (hd < tl) {
        int u = Q[++hd];
        for (rg int i = head[u]; i; i = edge[i].nxt) {
            int v = edge[i].v, w = edge[i].w;
            if (dep[v] == 0 && w)
                dep[v] = dep[u] + 1, Q[++tl] = v;
        }
    }
    return dep[t] > 0;
}

inline int dfs(int u, int flow) {
    if (u == t) return flow;
    for (rg int& i = cur[u]; i; i = edge[i].nxt) {
        int v = edge[i].v, w = edge[i].w;
        if (dep[v] == dep[u] + 1 && w) {
            int res = dfs(v, min(w, flow));
            if (res) { edge[i].w -= res, edge[i ^ 1].w += res; return res; }
        }
    }
    return 0;
}

inline int Dinic() {
    int res = 0;
    while (bfs()) {
        for (rg int i = s; i <= t; ++i) cur[i] = head[i];
        while (int d = dfs(s, INF)) res += d;
    }
    return res;
}

inline int id(int i, int j) { return (i - 1) * m + j; }

int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n), read(m), s = 0, t = n * m + 1;
    for (rg int i = 1; i <= n; ++i)
        for (rg int j = 1; j <= m; ++j) {
            read(x), sum += x > 0 ? x : -x;
            if (x > 0) link(s, id(i, j), x);
            if (x < 0) link(id(i, j), t, -x);
        }
    for (rg int i = 1; i <= n - 1; ++i)
        for (rg int j = 1; j <= m; ++j)
            read(x), link(id(i, j), id(i + 1, j), x), link(id(i + 1, j), id(i, j), x);
    for (rg int i = 1; i <= n; ++i)
        for (rg int j = 1; j <= m - 1; ++j)
            read(x), link(id(i, j), id(i, j + 1), x), link(id(i, j + 1), id(i, j), x);
    printf("%d\n", sum - Dinic());
    return 0;
}

原文地址：https://www.cnblogs.com/zsbzsb/p/12305485.html