HDU - 3035 War（對偶圖求最小割+最短路）

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題目連結:http://acm.hdu.edu.cn/showproblem.php?pid=3035

題意

給個圖，求把s和t分開的最小割。

分析

實際頂點和邊非常多，不能用最大流來求解。這道題要用平面圖求最小割的方法：

把面變成頂點，對每兩個面相鄰的邊作一條新邊。然後求最短路就是最小割了。

另外，外平面分成兩個點，分別是源點和匯點，源點連左下的邊，匯點連右上的邊，這樣跑出來才是正確的。

建圖參考自：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;}

 

HDU - 3035 War（對偶圖求最小割+最短路）