poj9125+2135 最小费用最大流<模版>

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the reason of falure:1、找最小min1的时候应该为cuo[pre[i]][i]-flow[pre[i]][i]而不是cup[pre[i]][i]

#include <iostream>#include <stdio.h>#include <queue>#include <string.h>#include <stdlib.h>using namespace std;struct ttt{int x,y;};int qq[500];int cost[500][500];int cup[500][500];int flow[500][500];int dist[500];ttt num1[500];ttt num2[500];int pre[500];int end1,start1;char map1[500][500];int bfs(int a,int b){    int i;    //cout <<"a="<< a << "b= "<<b << endl;    memset(pre,0,sizeof(pre));    for(i=a+1;i<=b;i++){        dist[i]=9999;    }    dist[a]=0;    queue<int>q1;    q1.push(a);    int u;    while(!q1.empty()){        u=q1.front();q1.pop();        for(i=a;i<=b;i++){            if(dist[i]>dist[u]+cost[u][i]&&cup[u][i]>flow[u][i]){                dist[i]=dist[u]+cost[u][i];                pre[i]=u;                //cout << i <<"的前面是" <<u << endl;                q1.push(i);            }        }    }   // cout << dist[b] <<endl;    if(dist[b]>=9999)return false;    return true;}int mincost(int a,int b){    int min1=9999,i;    int sum=0;    while(bfs(a,b)){        for(i=b;i!=a;i=pre[i]){            min1=min(min1,cup[pre[i]][i]-flow[pre[i]][i]);        }        for(i=b;i!=a;i=pre[i]){            flow[pre[i]][i]+=min1; //减去容量表面不能走了            flow[i][pre[i]]-=min1; //反向加容量说明可以走两次，多出来那次就相当与之前那次是没有走到的        }      //  cout << min1 <<endl;        sum+=min1*dist[b];    }    return sum;}int main(){    freopen("in.txt","r",stdin);    int t1,t2,t3,f1,f2,f3,i,j,k,l,n,m;    int k1,k2;    int g1,g2;    while(scanf("%d%d",&k1,&k2)&&(k1||k2)){        memset(map1,0,sizeof(map1));        memset(cost,0,sizeof(cost));        memset(num1,0,sizeof(num1));        memset(flow,0,sizeof(flow));        memset(num2,0,sizeof(num2));        memset(cup,0,sizeof(cup));        g1=0;g2=0;        for(i=1;i<=k1;i++){            scanf("%s",map1[i]);            for(j=0;j<k2;j++)            if(map1[i][j]=='m'){                g1++;                num1[g1].x=i;                num1[g1].y=j+1;            }else if(map1[i][j]=='H'){                g2++;                num2[g2].x=i;                num2[g2].y=j+1;            }        }            end1=g1+g2+1;            start1=0;            for(i=1;i<=g1;i++){ //cup为从0开始到任意一个人在的点容量为1，cost为0                cup[0][i]=1;                cost[0][i]=cost[i][0]=0;            }            for(j=1;j<=g2;j++){                cup[g1+j][end1]=1;                cost[g1+j][end1]=cost[end1][g1+j]=0;            }            for(i=1;i<=g1;i++)            for(j=1;j<=g2;j++){                cost[i][g1+j]=abs(num1[i].x-num2[j].x)+abs(num1[i].y-num2[j].y);                cost[g1+j][i]=-cost[i][g1+j];  //表明反方向可以走，只不过负数的这么一走，想想最大流反向走会怎么样                cup[i][g1+j]=1;            }            cout << mincost(start1,end1) <<endl;    }}

poj 2135

传纸条一样的意思。AC了，对于无向边，也就是可以走回来的，要连反向流量

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#include <queue>#define V 10100#define E 1000100#define inf 99999999using namespace std;int vis[V]; //V为点的数量 int dist[V];int pre[V];struct Edge{    int u,v,c,cost,next;}edge[E];int head[V],cnt;void init(){    cnt=0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int c,int cost){    edge[cnt].u=u;edge[cnt].v=v;edge[cnt].cost=cost;    edge[cnt].c=c;edge[cnt].next=head[u];head[u]=cnt++;    edge[cnt].u=v;edge[cnt].v=u;edge[cnt].cost=-cost;    edge[cnt].c=0;edge[cnt].next=head[v];head[v]=cnt++;}bool spfa(int begin,int end){    int u,v;    queue<int> q;    for(int i=0;i<=end+2;i++){        pre[i]=-1;        vis[i]=0;        dist[i]=inf;    }    vis[begin]=1;    dist[begin]=0;    q.push(begin);    while(!q.empty()){        u=q.front();        q.pop();        vis[u]=0;        for(int i=head[u];i!=-1;i=edge[i].next){            if(edge[i].c>0){                v=edge[i].v;                if(dist[v]>dist[u]+edge[i].cost){                    dist[v]=dist[u]+edge[i].cost;                    pre[v]=i;                    if(!vis[v]){                        vis[v]=true;                        q.push(v);                    }                }            }        }    }    return dist[end]!=inf;}int MCMF(int begin,int end){    int ans=0,flow;    int flow_sum=0;    while(spfa(begin,end)){        flow=inf;        for(int i=pre[end];i!=-1;i=pre[edge[i].u])            if(edge[i].c<flow)                flow=edge[i].c;        for(int i=pre[end];i!=-1;i=pre[edge[i].u]){            edge[i].c-=flow;            edge[i^1].c+=flow;        }        ans+=dist[end];        flow_sum += flow;    }    //cout << flow_sum << endl;    return ans;}int main(){    //freopen("in.txt","r",stdin);    int n,m,a,b,c,t1,t2,t3;    while(scanf("%d%d",&n,&m)!=EOF){        init();        addedge(0,1,2,0);        addedge(n,n+1,2,0);        for(int i=1;i<=m;i++){        scanf("%d %d %d",&t1,&t2,&t3);        addedge(t1,t2,1,t3); //最小费用最大流，无向边的话就创建2条addedge(t2,t1,1,t3); }cout <<MCMF(0,n+1) << endl;    }    return 0;}

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