HDU - 3035 War（对偶图求最小割+最短路）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3035

题意

给个图，求把s和t分开的最小割。

分析

实际顶点和边非常多，不能用最大流来求解。这道题要用平面图求最小割的方法：

把面变成顶点，对每两个面相邻的边作一条新边。然后求最短路就是最小割了。

另外，外平面分成两个点，分别是源点和汇点，源点连左下的边，汇点连右上的边，这样跑出来才是正确的。

建图参考自：https://blog.csdn.net/accelerator_/article/details/40957675

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

const int MAXNODE = 1000005;
const int MAXEDGE = 3 * MAXNODE;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
    int u, v;
    Type dist;
    Edge() {}
    Edge(int u, int v, Type dist) {
        this->u = u;
        this->v = v;
        this->dist = dist;
    }
};

struct HeapNode {
    Type d;
    int u;
    HeapNode() {}
    HeapNode(Type d, int u) {
        this->d = d;
        this->u = u;
    }
    bool operator < (const HeapNode& c) const {
        return d > c.d;
    }
};

struct Dijkstra {
    int n, m;
    Edge edges[MAXEDGE];
    int first[MAXNODE];
    int next[MAXEDGE];
    bool done[MAXNODE];
    Type d[MAXNODE];

    void init(int n) {
        this->n = n;
        memset(first, -1, sizeof(first));
        m = 0;
    }

    void add_Edge(int u, int v, Type dist) {
        edges[m] = Edge(u, v, dist);
        next[m] = first[u];
        first[u] = m++;
    }

    Type dijkstra(int s, int t) {
        priority_queue<HeapNode> Q;
        for (int i = 0; i < n; i++) d[i] = INF;
        d[s] = 0;
        memset(done, false, sizeof(done));
        Q.push(HeapNode(0, s));
        while (!Q.empty()) {
            HeapNode x = Q.top(); Q.pop();
            int u = x.u;
            if (done[u]) continue;
            done[u] = true;
            for (int i = first[u]; i != -1; i = next[i]) {
                Edge& e = edges[i];
                if (d[e.v] > d[u] + e.dist) {
                    d[e.v] = d[u] + e.dist;
                    Q.push(HeapNode(d[e.v], e.v));
                }
            }
        }
        return d[t];
    }
} gao;

typedef long long ll;

int n, m;

int main() {
    while (~scanf("%d%d", &n, &m)) {
        int u, v, w;
        gao.init(n * m * 4 + 2);
        int s = n * m * 4, t = n * m * 4 + 1;
        for (int i = 0; i < (n + 1); i++) {
            for (int j = 0; j < m; j++) {
                scanf("%d", &w);
                u = (i - 1) * m + j + n * m;
                v = i * m + j;
                if (i == 0) u = t;
                if (i == n) v = s;
                gao.add_Edge(u, v, w);
                gao.add_Edge(v, u, w);
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < (m + 1); j++) {
                scanf("%d", &w);
                u = n * m * 3 + i * m + j - 1;
                v = n * m * 2 + i * m + j;
                if (j == 0) u = s;
                if (j == m) v = t;
                gao.add_Edge(u, v, w);
                gao.add_Edge(v, u, w);
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                scanf("%d", &w);
                u = i * m + j;
                v = n * m * 2 + i * m + j;
                gao.add_Edge(u, v, w);
                gao.add_Edge(v, u, w);
                scanf("%d", &w);
                v += n * m;
                gao.add_Edge(u, v, w);
                gao.add_Edge(v, u, w);
            }
            for (int j = 0; j < m; j++) {
                scanf("%d", &w);
                u = n * m + i * m + j;
                v = n * m * 2 + i * m + j;
                gao.add_Edge(u, v, w);
                gao.add_Edge(v, u, w);
                scanf("%d", &w);
                v += n * m;
                gao.add_Edge(u, v, w);
                gao.add_Edge(v, u, w);
            }
        }
        printf("%d\n", gao.dijkstra(s, t));
    }
    return 0;
}

原文地址：https://www.cnblogs.com/fht-litost/p/9773497.html