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题意：
AP x y
A recruit with ability rate x were asked to join company y. (0<=x<2^31, 0<=y<n)
MG x y
Company x and company y is merged. The new company is numbered as x. (0<=x, y<n)
GT x
Report the fighting capacity of company x. (0<=x<n)
思路：
并查集，判断好各种可能的情况。
代码：
int n,k,m; int fa[100005], ability[100005]; int findFa(int x){ if(fa[x]==x) return fa[x]; int t=fa[x]; fa[x]=findFa(fa[x]); ability[x] = min( ability[x],ability[t] ); return fa[x]; } int main(){ while(scanf("%d%d%d",&n,&k,&m)!=EOF){ rep(i,0,n-1){ fa[i]=i; ability[i]=inf; } rep(i,1,k){ int rate,belongTo; scanf("%d%d",&rate,&belongTo); ability[belongTo]=min( ability[belongTo],rate ); } while(m--){ char ope[5]; scanf("%s",ope); if(ope[0]==‘A‘){ int x,y; scanf("%d%d",&x,&y); int faTemp=findFa(y); if(y!=faTemp){ puts("Reject"); }else{ ability[faTemp]=min( ability[faTemp],x ); puts("Accept"); } } else if(ope[0]==‘M‘){ int x,y; scanf("%d%d",&x,&y); int fx=findFa(x); int fy=findFa(y); if(fx!=x || fy!=y || fx==fy){ puts("Reject"); }else{ ability[fx]=min( ability[fx],ability[fy] ); fa[fy]=fx; puts("Accept"); } } else{ int x; scanf("%d",&x); int fx=findFa(x); if(x!=fx){ printf("Company %d is a part of company %d.\n",x,fx); } else{ if(ability[x]==inf){ printf("Company %d is empty.\n",x); }else{ printf("Lowest rate: %d.\n",ability[x]); } } } } puts(""); } return 0; }
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原文地址：http://www.cnblogs.com/fish7/p/4332537.html
