POJ 2421Constructing Roads

题意：提供任意两点直接的距离，然后在已知某两点直接已经修建好路的情况下，问还需要修长的最小距离，注意！再用 kruskal 算法的时候注意有些已经修好的边可能出现重复的  所以需要判断处理，其他都是套路，细心处理即可；

#include<algorithm>#include<iostream>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include<cstring>#include<string>using namespace std;int a[105][105];int first[105];int next[105];typedef struct node{int x;int y;int date;}node;bool cmp(node w,node e){return w.date<e.date;}node vulue[10005];void csh(int m){for(int i=1;i<=m;i++){first[i]=i;}for(int i=1;i<=m;i++){next[i]=1;}}int find(int x){if(first[x]==x)  return x;else{x=first[x];return find(x);}}void hebin(int x,int y){x=find(x);y=find(y);if(x==y) return ;if(next[x]>next[y]){first[y]=x;}else{first[x]=y;if(next[x]==next[y]){next[y]++;}}return ;}int main(){int n;cin>>n;csh(n);int k=0;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){cin>>a[i][j];if(a[i][j]!=0&&i<=j){k++;vulue[k].x=i;vulue[k].y=j;vulue[k].date=a[i][j];}}}sort(vulue+1,vulue+k+1,cmp);int m;cin>>m;int b,c;int sum=0;for(int i=1;i<=m;i++){cin>>b>>c;if(find(b)!=find(c)){sum++;hebin(b,c);}}int ans=0;for(int i=1;i<=k;i++){if(find(vulue[i].x)!=find(vulue[i].y)){sum++;ans+=vulue[i].date;hebin(vulue[i].x,vulue[i].y);if(sum==(n-1)){break;}}}cout<<ans<<endl;return 0;}