最小费用最大流模板

最小费用最大流一般用邻接表来实现，因为邻接矩阵不能处理平行边等等；而一条有向边是要储存两条信息，无向图的话要拆成两条有向边处理，相当于变为4条边，这也是邻接矩阵不能做到的

然后最小费用最大流的原理就不讲了，讲一下实现的要注意的问题和一些技巧

1.用结构体数组来保存边，最好从下标0开始保存不要从下标1开始保存，因为增广的时候需要用到位运算，从下标1保存不利于位运算

2.记录路径：如果是邻接矩阵记录前驱p[v]=u，表示一条边u—>v,u就是v的前驱

所以找出整条路径就是一个循环 for(u=t; u!=s; u=p[u])

邻接表p[v]=i; //i表示的是第i条边，那么它的前驱应该是e[i].u,即在这条边的信息内部

所以找出整条路径是 for(i=p[t]; i!=-1; i=p[e[i].u]) //i表示的是边的标号

3.最小费用最大流：

I.在还能增流的条件下，以单位费用作为权去求源点到汇点最短路并且需要记录路径，但记住不要改变流量，改变流量是增广的时候做的，因为单位费用可能有负值所以用spfa。(spfa算法)

II.如果能得到到汇点的最短路，说明还没到达最小费用最大流，那么就沿刚才记录的路径返回，找出最小残余流量

III.增广：即再沿路径返回一次，添加残余流量，反向则减少残余流量

IV.更新最大流，和最小费用 即F+=min; C+=d[t]*min; //min是本次的最小参与流量

#define MAXN 1003#define MAXM 40004using namespace std; const int INF = 0x3f3f3f3f;struct EDGE{    int to, cap, next, flow, cost;}edge[MAXM];int q[MAXN], head[MAXN], tol;int pre[MAXN], dis[MAXN];bool vis[MAXN];int n, m;void init(){    tol = 0;    memset(head, -1, sizeof(head));}void addEdge(int u, int v, int cap, int cost){    edge[tol].cap = cap;    edge[tol].cost = cost;    edge[tol].flow = 0;    edge[tol].next = head[u];    edge[tol].to = v;    head[u] = tol++;    edge[tol].cap = 0;    edge[tol].cost = -cost;    edge[tol].flow = 0;    edge[tol].next = head[v];    edge[tol].to = u;    head[v] = tol++;}bool spfa(int st, int en){    int rear = 0, front = 0;    for (int i = st; i <= en+1; i++){        dis[i] = INF;        vis[i] = false;        pre[i] = -1;    }    dis[st] = 0;    vis[st] = true;    q[front++] = st;    while(rear < front){        int u = q[rear++];        vis[u] = false;        for (int e = head[u]; e != -1; e = edge[e].next){            int v = edge[e].to;            if (edge[e].cap > edge[e].flow && dis[v] > dis[u]+edge[e].cost){                dis[v] = dis[u] + edge[e].cost;                pre[v] = e;                if (!vis[v]){                    vis[v] = true;                    q[front++] = v;                }            }        }    }    return pre[en] != -1;}int minCostMaxflow(int st, int en, int &cost){    int flow = 0;    while(spfa(st, en)){        int Min = INF;        for (int i = pre[en]; i != -1; i = pre[edge[i^1].to]){            Min = Min > (edge[i].cap-edge[i].flow) ? (edge[i].cap-edge[i].flow) : Min;        }        for (int i = pre[en]; i != -1; i = pre[edge[i^1].to]){            edge[i].flow += Min;            edge[i^1].flow -= Min;            cost += Min*edge[i].cost;         }    }    return flow;}

---

poj 2135
最小费用最大流模板题

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define MAXN 1003#define MAXM 40004using namespace std; const int INF = 0x3f3f3f3f;struct EDGE{    int to, cap, next, flow, cost;}edge[MAXM];int q[1000000], head[MAXN], tol;int pre[MAXN], dis[MAXN];bool vis[MAXN];int n, m;void init(){    tol = 0;    memset(head, -1, sizeof(head));}void addEdge(int u, int v, int cap, int cost){    edge[tol].cap = cap;    edge[tol].cost = cost;    edge[tol].flow = 0;    edge[tol].next = head[u];    edge[tol].to = v;    head[u] = tol++;    edge[tol].cap = 0;    edge[tol].cost = -cost;    edge[tol].flow = 0;    edge[tol].next = head[v];    edge[tol].to = u;    head[v] = tol++;}bool spfa(int st, int en){    int rear = 0, front = 0;    for (int i = st; i <= en+1; i++){        dis[i] = INF;        vis[i] = false;        pre[i] = -1;    }    dis[st] = 0;    vis[st] = true;    q[front++] = st;    while(rear < front){        int u = q[rear++];        vis[u] = false;        for (int e = head[u]; e != -1; e = edge[e].next){            int v = edge[e].to;            if (edge[e].cap > edge[e].flow && dis[v] > dis[u]+edge[e].cost){                dis[v] = dis[u] + edge[e].cost;                pre[v] = e;                if (!vis[v]){                    vis[v] = true;                    q[front++] = v;                }            }        }    }    return pre[en] != -1;}int minCostMaxflow(int st, int en, int &cost){    int flow = 0;    while(spfa(st, en)){        int Min = INF;        for (int i = pre[en]; i != -1; i = pre[edge[i^1].to]){            Min = Min > (edge[i].cap-edge[i].flow) ? (edge[i].cap-edge[i].flow) : Min;        }        for (int i = pre[en]; i != -1; i = pre[edge[i^1].to]){            edge[i].flow += Min;            edge[i^1].flow -= Min;        }        cost += dis[en] * Min;    }    return flow;}int main(){    int i, j, k, cost, u, v;    scanf("%d%d", &n, &m);    init();    for (i = 0; i < m; i++){        scanf("%d%d%d", &u, &v, &cost);        addEdge(u, v, 1, cost);        addEdge(v, u, 1, cost);    }    addEdge(0, 1, 2, 0);    addEdge(n, n+1, 2, 0);    cost = 0;    minCostMaxflow(0, n+1, cost);    printf("%d\n", cost);    return 0;}