标签:animal tput sts nal div tran otto nes scan
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1332 Accepted Submission(s): 664
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int T,n,m,map[20][20],fa[20],a[20],b[20],ans; int main(){ scanf("%d",&T); while(T--){ memset(map,0,sizeof(map)); scanf("%d%d",&n,&m); for(int i=1;i<=m;i++){ int x,y; scanf("%d%d",&x,&y); x+=1;y+=1; map[x][y]++; map[y][x]++; } ans=0x7f7f7f7f; for(int i=0;i<(1<<n);i++){//n个点分成两边,有2^n种分法。 int maxn=0; int sa=0,sb=0; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); for(int j=1;j<=n;j++) if(i&(1<<(j-1))) a[++sa]=j; else b[++sb]=j; for(int j=1;j<=sa;j++) for(int k=j+1;k<=sa;k++) if(map[a[j]][a[k]]) maxn+=map[a[j]][a[k]]; for(int j=1;j<=sb;j++) for(int k=j+1;k<=sb;k++) if(map[b[j]][b[k]]) maxn+=map[b[j]][b[k]]; ans=min(ans,maxn); } printf("%d\n",ans); } }
标签:animal tput sts nal div tran otto nes scan
原文地址:http://www.cnblogs.com/cangT-Tlan/p/7436428.html