hdu 1853 | hdu 3488 | hdu3435

 題目：http://acm.hdu.edu.cn/showproblem.php?pid=1853

二分圖的最優匹配。求最小費用流。要求所有點匹配下的最小費用，直接套用KM演算法的即可。。

下面是 1853 AC代碼：

#include<cstdio>#include<cstring>using namespace std;const int maxn = 305;const int INF = (1<<30)-1;int g[maxn][maxn];int lx[maxn],ly[maxn];int match[maxn];bool visx[maxn],visy[maxn];int slack[maxn];int n;bool dfs(int cur){     visx[cur] = true;     for(int y=1;y<=n;y++){         if(visy[y])   continue;         int t=lx[cur]+ly[y]-g[cur][y];         if(t==0){            visy[y] = true;            if(match[y]==-1||dfs(match[y])){                match[y] = cur;                return true;            }         }         else if(slack[y]>t){                 slack[y]=t;         }     }     return false;}int KM(){    memset(match,-1,sizeof(match));    memset(ly,0,sizeof(ly));    for(int i=1 ;i<=n;i++){         lx[i]=-INF;       for(int j=1;j<=n;j++)           if(g[i][j]>lx[i])   lx[i]=g[i][j];   }   for(int x=1;x<=n;x++){        for(int i=1;i<=n;i++)  slack[i]=INF;        while(true){            memset(visx,false,sizeof(visx));            memset(visy,false,sizeof(visy));            if(dfs(x))  break;            int d=INF;            for(int i=1;i<=n;i++){               if(!visy[i]&&d>slack[i])     d=slack[i];            }            for(int i=1;i<=n;i++){               if(visx[i])                  lx[i]-=d;            }            for(int i=1;i<=n;i++){               if(visy[i])                 ly[i]+=d;               else                        slack[i]-=d;            }        }   }    int result = 0;    for(int i = 1; i <=n; i++){        if(match[i]==-1||g[match[i]][i]==-INF)            return 1;      if(match[i]>-1)        result += g[match[i]][i];    }    return result;}int main(){    int m;    int a,b,c;    while(scanf("%d%d",&n,&m)!=EOF){        memset(g,0,sizeof(g));        for(int i=1;i<=n;i++)         for(int j=1;j<=n;j++)             g[i][j]=-INF;        for(int i=0;i<m;i++) {            scanf("%d%d%d",&a,&b,&c);            if(g[a][b]<-c)            g[a][b]=-c;        }           int ans=-KM();           printf("%d\n",ans);    }    return 0;}

下面是3488AC代碼：

#include<cstdio>#include<cstring>using namespace std;const int maxn = 205;const int INF = (1<<30)-1;int g[maxn][maxn];int lx[maxn],ly[maxn];int match[maxn];bool visx[maxn],visy[maxn];int slack[maxn];int n;bool dfs(int cur){     visx[cur] = true;     for(int y=1;y<=n;y++){         if(visy[y])   continue;         int t=lx[cur]+ly[y]-g[cur][y];         if(t==0){            visy[y] = true;            if(match[y]==-1||dfs(match[y])){                match[y] = cur;                return true;            }         }         else if(slack[y]>t){                 slack[y]=t;         }     }     return false;}int KM(){    memset(match,-1,sizeof(match));    memset(ly,0,sizeof(ly));    for(int i=1 ;i<=n;i++){         lx[i]=-INF;       for(int j=1;j<=n;j++)           if(g[i][j]>lx[i])   lx[i]=g[i][j];   }   for(int x=1;x<=n;x++){        for(int i=1;i<=n;i++)  slack[i]=INF;        while(true){            memset(visx,false,sizeof(visx));            memset(visy,false,sizeof(visy));            if(dfs(x))  break;            int d=INF;            for(int i=1;i<=n;i++){               if(!visy[i]&&d>slack[i])     d=slack[i];            }            for(int i=1;i<=n;i++){               if(visx[i])                  lx[i]-=d;            }            for(int i=1;i<=n;i++){               if(visy[i])                 ly[i]+=d;               else                        slack[i]-=d;            }        }   }    int result = 0;    for(int i = 1; i <=n; i++){        if(match[i]==-1||g[match[i]][i]==-INF)            return 1;      if(match[i]>-1)        result += g[match[i]][i];    }    return result;}int main(){    int m,T;    int a,b,c;    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        memset(g,0,sizeof(g));        for(int i=1;i<=n;i++)         for(int j=1;j<=n;j++)             g[i][j]=-INF;        for(int i=0;i<m;i++) {            scanf("%d%d%d",&a,&b,&c);            if(g[a][b]<-c)            g[a][b]=-c;        }           int ans=-KM();           printf("%d\n",ans);    }    return 0;}

下面是 3435 AC 代碼：

#include<cstdio>#include<cstring>using namespace std;const int maxn = 1005;const int INF = (1<<30)-1;int g[maxn][maxn];int lx[maxn],ly[maxn];int match[maxn];bool visx[maxn],visy[maxn];int slack[maxn];int n;bool dfs(int cur){     visx[cur] = true;     for(int y=1;y<=n;y++){         if(visy[y])   continue;         int t=lx[cur]+ly[y]-g[cur][y];         if(t==0){            visy[y] = true;            if(match[y]==-1||dfs(match[y])){                match[y] = cur;                return true;            }         }         else if(slack[y]>t){                 slack[y]=t;         }     }     return false;}int KM(){    memset(match,-1,sizeof(match));    memset(ly,0,sizeof(ly));    for(int i=1 ;i<=n;i++){         lx[i]=-INF;       for(int j=1;j<=n;j++)           if(g[i][j]>lx[i])   lx[i]=g[i][j];   }   for(int x=1;x<=n;x++){        for(int i=1;i<=n;i++)  slack[i]=INF;        while(true){            memset(visx,false,sizeof(visx));            memset(visy,false,sizeof(visy));            if(dfs(x))  break;            int d=INF;            for(int i=1;i<=n;i++){               if(!visy[i]&&d>slack[i])     d=slack[i];            }            for(int i=1;i<=n;i++){               if(visx[i])                  lx[i]-=d;            }            for(int i=1;i<=n;i++){               if(visy[i])                 ly[i]+=d;               else                        slack[i]-=d;            }        }   }    int result = 0;    for(int i = 1; i <=n; i++){        if(match[i]==-1||g[match[i]][i]==-INF)            return 1;      if(match[i]>-1)        result += g[match[i]][i];    }    return result;}int main(){    int m,T,ca=1;    int a,b,c;    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        memset(g,0,sizeof(g));        for(int i=1;i<=n;i++)         for(int j=1;j<=n;j++)             g[i][j]=-INF;        for(int i=0;i<m;i++) {            scanf("%d%d%d",&a,&b,&c);            if(g[a][b]<-c)            g[a][b]=-c;            if(g[b][a]<-c)            g[b][a]=-c;        }           int ans=-KM();        if(ans==-1)           printf("Case %d: NO\n",ca++);        else           printf("Case %d: %d\n",ca++,ans);    }    return 0;}