【最小費用最大流】POJ 2195

可以KM，也可以費用流，最近學費用流，所以用一下，debug真浪費時間，以為spfa有問題，原來是計數時出錯。。。囧啊，菜啊，話說poj 2516到現在還不知道wa哪裡，抓狂啊！！！

#define inf 1<<30#define N 222int cap[N][N];int cost[N][N];int pre[N];int dis[N];bool vis[N];int minc;char str[110][110];int min(int a,int b){return a>b?b:a;}void spfa(int s,int t){    queue<int> qq;    int i,j;    while(1){        for(i=s;i<=t;i++){            vis[i] = 0;            dis[i] = inf;        }        dis[s] = 0;        qq.push(s);        vis[s] = 1;        while(!qq.empty()){            int u = qq.front();            qq.pop();            vis[u] = 0;            for(int v=s;v<=t;v++){                if(cap[u][v] && dis[u] + cost[u][v] < dis[v]){                    dis[v] = dis[u] + cost[u][v];                    pre[v] = u;                    if(!vis[v]){                        vis[v] = 1;                        qq.push(v);                    }                }            }        }        if(dis[t] == inf){            return ;        }        int a = inf;        for(int u=t;u!=s;u=pre[u]){            a = min(a,cap[pre[u]][u]);        }        for(int u=t;u!=s;u=pre[u]){            cap[pre[u]][u] -= a;            cap[u][pre[u]] += a;        }        minc += dis[t]*a;    }    return ;}struct node{    int r,c;}mm[N/2],hh[N/2];int cntm ;int cnth ;void work(){    int i,j;    int s = 0,t = cntm + cnth + 1;    minc =0;    memset(cap,0,sizeof(cap));    memset(cost,0,sizeof(cost));    for(i=1;i<=cntm;i++){        cap[s][i] = 1;//每條邊容量為1，最後肯定是滿流的        cost[s][i] = 0;    }    for(i=1;i<=cnth;i++){        cap[i+cntm][t] = 1;        cost[i+cntm][t] = 0;    }    for(i=1;i<=cntm;i++){        for(j=1;j<=cnth;j++){            cap[i][j+cntm] = 1;            cost[i][j+cntm] = abs(mm[i].r - hh[j].r) + abs(mm[i].c - hh[j].c);            cost[j+cntm][i] = -cost[i][j+cntm];//注意要取反        }    }    spfa(s,t);    printf("%d\n",minc);}int main(){    int n,m;    while(scanf("%d%d",&n,&m) && (n+m)){        int i,j;        for(i=1;i<=n;i++){            scanf("%s",str[i]+1);        }        cntm = 0;        cnth = 0;        for(i=1;i<=n;i++){            for(j=1;j<=m;j++){                if(str[i][j] == 'm'){                    mm[++cntm].r = i;                    mm[cntm].c = j;                } else if(str[i][j] == 'H'){                    hh[++cnth].r = i;                    hh[cnth].c = j;                }            }        }cout<<cntm<<endl;        work();    }    return 0;}