最小费用最大流（板子）

来源：互联网发布：网络直播举报编辑：程序博客网时间：2024/06/06 13:05

写最小费用最大流的时候比较容易忘的是添加一条边的费用时忘记将反向边的费用置为原来边的相反数。

这一点要特别注意。

邻接矩阵版本

//************************************************************//最小费用最大流算法//SPFA求最短路//邻接矩阵形式//初始化:cap:容量，没有边为0//cost:耗费，对称形式，没有边的也为0//c是最小费用//f是最大流//*******************************************************const int MAXN=500;const int INF=0x3fffffff;int cap[MAXN][MAXN];//容量，没有边为0int flow[MAXN][MAXN];//耗费矩阵是对称的，有i到j的费用，则j到i的费用为其相反数int cost[MAXN][MAXN];int n;//顶点数目0~n-1int f;//最大流int c;//最小费用int start,end;//源点和汇点bool vis[MAXN];//在队列标志int que[MAXN];int pre[MAXN];int dist[MAXN];//s-t路径最小耗费bool SPFA(){    int front=0,rear=0;    for(int u=0;u<=n;u++)    {        if(u==start)        {            que[rear++]=u;            dist[u]=0;            vis[u]=true;        }        else        {            dist[u]=INF;            vis[u]=false;        }    }    while(front!=rear)    {        int u=que[front++];        vis[u]=false;        if(front>=MAXN)front=0;        for(int v=0;v<=n;v++)        {            if(cap[u][v]>flow[u][v]&&dist[v]>dist[u]+cost[u][v])            {                dist[v]=dist[u]+cost[u][v];                pre[v]=u;                if(!vis[v])                {                    vis[v]=true;                    que[rear++]=v;                    if(rear>=MAXN)rear=0;                }            }        }    }    if(dist[end]>=INF)return false;    return true;}void minCostMaxflow(){    memset(flow,0,sizeof(flow));    c=f=0;    while(SPFA())    {        int Min=INF;        for(int u=end;u!=start;u=pre[u])           Min=min(Min,cap[pre[u]][u]-flow[pre[u]][u]);        for(int u=end;u!=start;u=pre[u])        {            flow[pre[u]][u]+=Min;            flow[u][pre[u]]-=Min;        }        c+=dist[end]*Min;        f+=Min;    }}

邻接表版本

#include <stdio.h>#include <algorithm>#include <string.h>#include <iostream>#include <string>#include <queue>using namespace std;const int MAXN = 10000;const int MAXM = 100000;const int INF = 0x3f3f3f3f;struct Edge{    int to,next,cap,flow,cost;}edge[MAXM];int head[MAXN],tol;int pre[MAXN],dis[MAXN];bool vis[MAXN];int N;//节点总个数，节点编号从0~N-1void init(int n){    N = n;    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;    for(int i = 0;i < N;i++)    {        dis[i] = INF;        vis[i] = false;        pre[i] = -1;    }    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);                }            }        }    }    if(pre[t] == -1)return false;    else return true;}//返回的是最大流，cost存的是最小费用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;}

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