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 = ;
const int MAXEDGE = * 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, -, sizeof(first));
m = ;
} 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 = ; i < n; i++) d[i] = INF;
d[s] = ;
memset(done, false, sizeof(done));
Q.push(HeapNode(, 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 != -; 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 * + );
int s = n * m * , t = n * m * + ;
for (int i = ; i < (n + ); i++) {
for (int j = ; j < m; j++) {
scanf("%d", &w);
u = (i - ) * m + j + n * m;
v = i * m + j;
if (i == ) u = t;
if (i == n) v = s;
gao.add_Edge(u, v, w);
gao.add_Edge(v, u, w);
}
}
for (int i = ; i < n; i++) {
for (int j = ; j < (m + ); j++) {
scanf("%d", &w);
u = n * m * + i * m + j - ;
v = n * m * + i * m + j;
if (j == ) u = s;
if (j == m) v = t;
gao.add_Edge(u, v, w);
gao.add_Edge(v, u, w);
}
}
for (int i = ; i < n; i++) {
for (int j = ; j < m; j++) {
scanf("%d", &w);
u = i * m + j;
v = n * m * + 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 = ; j < m; j++) {
scanf("%d", &w);
u = n * m + i * m + j;
v = n * m * + 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 ;
}
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