题目:http://acm.hdu.edu.cn/showproblem.php?pid=1853
二分图的最优匹配。求最小费用流。要求所有点匹配下的最小费用,直接套用KM算法的即可。。
下面是 1853 AC代码:
#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代码:
#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 代码:
#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; }