PTA 1034. Head of a Gang (30)

使用并查集

#include<stdio.h>#include<string.h>#include<string>#include<iostream>#include<queue>#include<stack>#include<math.h>#include<string>#include<algorithm>#include<map>using namespace std;vector<int>v[2];//保存输入的对话 int a[260000];//并查集 int value[260000];//每个人的weight int all[260000];//每个连通块的总weight 使用时除2  int find(int x){if(a[x] < 0)return x;return a[x] = find(a[x]);}void insert(int x,int y){int p = find(x);int q = find(y);if(p == q)return ;if(value[p] > value[q])a[q] = p;//保证连通块头点为最大weight else a[p] = q;}int main(){int n,m;scanf("%d%d",&n,&m);memset(value,0,sizeof(value));memset(all,0,sizeof(all));memset(a,-1,sizeof(a));for(int i=0;i<n;i++){//数字化，每一个字母对应两位数，直接用数组 char x[4],y[4];int p,q,t;scanf("%s",x);getchar();scanf("%s",y);scanf("%d",&t);p = (x[0]-'A')*10000+(x[1]-'A')*100+(x[2]-'A');q = (y[0]-'A')*10000+(y[1]-'A')*100+(y[2]-'A');value[p] += t;value[q] += t;v[0].push_back(p);v[1].push_back(q);}for(int i=0;i<n;i++){//并查集 insert(v[0][i],v[1][i]);}bool book[260000];//用来标记该点是否已经计算过 memset(book,false,sizeof(book));vector<int>r;//保存所有连通块的头结点 for(int i=0;i<n;i++){for(int j=0;j<2;j++){if(!book[v[j][i]]){book[v[j][i]] = 1;if(a[v[j][i]] >= 0){//不为头结点 int point = find(v[j][i]);//找到头节点 a[point]--;//用负值来保存该连通块共有几个点，初始值为-1 all[point] += value[v[j][i]];//合计连通块weight }else{//头结点 all[v[j][i]] += value[v[j][i]];r.push_back(v[j][i]);}}}}vector<int >real;//保存结果 for(int i=0;i<r.size();i++){if(a[r[i]] < -2 && all[r[i]]/2 > m)real.push_back(r[i]);//判断是否符合条件 }printf("%d\n",real.size());sort(real.begin(),real.end());for(int i=0;i<real.size();i++){//复原 int rea = real[i];char temp[4];temp[0] = rea/10000+'A';rea%=10000;temp[1] = rea/100+'A';rea%=100;temp[2] = rea+'A';temp[3] = 0;printf("%s %d\n",temp,-a[real[i]]);}return 0;}