POJ 2135 Farm Tour【最小费用流】

来源：互联网发布：中国加工贸易数据编辑：程序博客网时间：2024/06/05 01:10

题意：N个点，M条边，从1号点出发到N号点，再走回来，且不能走走过的路，求最小路径权和

最小费用流，用流量为1 限制每条边只能走一次，建立超源汇点，源点连1号点的流量为2，费用0，汇点同理。其他边的流量为1，费用为权值。

建立超源汇点且权值为2目的是因为起终点经过2次（虽然这里是点经过2次而题目是边），而如果不建超源点的话没办法表示，如果是直接把跟1号点相连的都设置为2，则很多边都可以走2次了，不科学了。

建立a->b的边时建反向弧，也建立b->a的边，这样当一去一回的情况下流量就为0了

下面模板是kuangbin的，模板大法好

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <set>#include <map>#include <stack>#include <queue>#include <vector>#include <iostream>#include <algorithm>using namespace std;#define ll long long#define eps 10^(-6)#define Q_CIN ios::sync_with_stdio(false)#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define CLR( a , x ) memset ( a , x , sizeof (a) )#define RE freopen("1.in","r",stdin);#define WE freopen("1.out","w",stdout);#define MOD 10009#define NMAX 10002//模板大法好const int inf = 0x3f3f3f3f;const int maxn = 1005;const int maxm = 10000 * 2 + 10;struct Edge {    int to, next, cap, flow, cost;} edge[maxm * 2];int head[maxn], tol;int pre[maxn], dis[maxn];bool vis[maxn];void init() {    tol = 0;    memset(head, -1, sizeof(head));}void addEdge(int u, int v, int cap, int cost) {    edge[tol].to = v;    edge[tol].cap = cap;    edge[tol].cost = cost;    edge[tol].flow = 0;    edge[tol].next = head[u];    head[u] = tol++;    edge[tol].to = u;    edge[tol].cap = 0;    edge[tol].cost = -cost;    edge[tol].flow = 0;    edge[tol].next = head[v];    head[v] = tol++;}bool spfa(int s, int t) {    queue<int>q;    memset(dis, inf, sizeof(dis));    memset(vis, false, sizeof(vis));    memset(pre, -1, sizeof(pre));    dis[s] = 0;    vis[s] = true;    q.push(s);    while (!q.empty()) {        int u = q.front();        q.pop();        vis[u] = false;        for (int i = head[u]; i != -1; i = edge[i].next) {            int v = edge[i].to;            if (edge[i].cap > edge[i].flow &&                    dis[v] > dis[u] + edge[i].cost ) {                dis[v] = dis[u] + edge[i].cost;                pre[v] = i;                if (!vis[v]) {                    vis[v] = true;                    q.push(v);                }            }        }    }    return pre[t] != -1;}//返回最大流，cost存最小费用int minCostMaxflow(int s, int t, int &cost) {    int flow = 0;    cost = 0;    while (spfa(s, t)) {        int Min = inf;        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {            if (Min > edge[i].cap - edge[i].flow) {                Min = edge[i].cap - edge[i].flow;            }        }        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {            edge[i].flow += Min;            edge[i ^ 1].flow -= Min;            cost += edge[i].cost * Min;        }        flow += Min;    }    return flow;}int main() {    int n, m, a, b, c;    //        freopen("1.in","r",stdin);    while (~scanf("%d%d",&n,&m)) {        int ss = 0, tt = n + 1;        init();        addEdge(ss, 1, 2, 0);        addEdge(n, tt, 2, 0);        while (m--) {            scanf("%d%d%d",&a,&b,&c);            addEdge(a, b, 1, c);            addEdge(b, a, 1, c);        }        int cost;        minCostMaxflow(ss, tt, cost);        printf("%d\n",cost);    }    return 0;}

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