POJ 2135 Farm Tour(费用流)

POJ 2135 Farm Tour

题目链接

题意：给定一个无向图，边有权值，求从1到n再从n到1的最短路

思路：费用流，连边容量为1（注意是无向图），然后源点和1连容量2，n和汇点连容量是2

代码：

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;const int MAXNODE = 1005;const int MAXEDGE = 40005;typedef long long Type;const Type INF = 0x3f3f3f3f3f3f3f;struct Edge {int u, v;Type cap, flow, cost;Edge() {}Edge(int u, int v, Type cap, Type flow, Type cost) {this->u = u;this->v = v;this->cap = cap;this->flow = flow;this->cost = cost;}};struct MCFC {int n, m, s, t;Edge edges[MAXEDGE];int first[MAXNODE];int next[MAXEDGE];int inq[MAXNODE];Type d[MAXNODE];int p[MAXNODE];Type a[MAXNODE];void init(int n) {this->n = n;memset(first, -1, sizeof(first));m = 0;}void add_Edge(int u, int v, Type cap, Type cost) {edges[m] = Edge(u, v, cap, 0, cost);next[m] = first[u];first[u] = m++;edges[m] = Edge(v, u, 0, 0, -cost);next[m] = first[v];first[v] = m++;}bool bellmanford(int s, int t, Type &flow, Type &cost) {for (int i = 0; i < n; i++) d[i] = INF;memset(inq, false, sizeof(inq));d[s] = 0; inq[s] = true; p[s] = s; a[s] = INF;queue<int> Q;Q.push(s);while (!Q.empty()) {int u = Q.front(); Q.pop();inq[u] = false;for (int i = first[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (e.cap > e.flow && d[e.v] > d[u] + e.cost) {d[e.v] = d[u] + e.cost;p[e.v] = i;a[e.v] = min(a[u], e.cap - e.flow);if (!inq[e.v]) {Q.push(e.v); inq[e.v] = true;}}}}if (d[t] == INF) return false;flow += a[t];cost += d[t] * a[t];int u = t;while (u != s) {edges[p[u]].flow += a[t];edges[p[u]^1].flow -= a[t];u = edges[p[u]].u;}return true;}Type Mincost(int s, int t) {Type flow = 0, cost = 0;while (bellmanford(s, t, flow, cost));return cost;}} gao;typedef long long ll;int n, m;int main() {while (~scanf("%d%d", &n, &m)) {gao.init(n + 2);int u, v;ll w;while (m--) {scanf("%d%d%lld", &u, &v, &w);gao.add_Edge(u, v, 1, w);gao.add_Edge(v, u, 1, w);}gao.add_Edge(0, 1, 2, 0);gao.add_Edge(n, n + 1, 2, 0);printf("%lld\n", gao.Mincost(0, n + 1));}return 0;}