[2016-03-19][UVALive][3027][Corporative Network]

• 时间:2016-03-19 13:24:23 星期六

• 题目编号:[2016-03-19][UVALive][3027][Corporative Network]

• 题目大意:给定n个节点,I u v表示把u节点的父节点设置为v,距离为|u-v|%1000,E u表示询问u到根节点的距离,给定若干个I E操作,输出相应答案

• 分析:带权并查集

• 方法:d[maxn] 维护到父节点的距离,每次询问的时候,把当前节点压缩到根节点输出对应距离即可

#ifdef _WORK_
#include <algorithm>
#include <cstdio>
#include<cstring>
using namespace std;
#define CLR(x,y) memset((x),(y),sizeof((x)))
#define FOR(x,y,z) for(int (x)=(y);(x)<(z);++(x))
const int maxn = 20000 +10;
int fa[maxn],d[maxn];
inline void ini(int n){
        CLR(d,0);
        FOR(i,0,n+1)    fa[i] = i;
}
int fnd(int x){
        if(x == fa[x])  return x;
        int tmp = fa[x];
        fa[x] = fnd(fa[x]);
        d[x] = d[x] + d[tmp];
        return fa[x];
}
int main(){
        int t,n,p,q;
        char str[10];
        scanf("%d",&t);
        while(t--){
                scanf("%d",&n);
                ini(n);
                while(~scanf("%s",str) && str[0] != ‘O‘){
                        if(str[0] == ‘I‘){
                                scanf("%d%d",&p,&q);
                                fa[p] = q;
                                d[p] = abs(p - q)%1000;
                        }else{
                                scanf("%d",&p);
                                fnd(p);
                                printf("%d\n",d[p]);
                        }
                }
        }
        return 0;
}

[2016-03-19][UVALive][3027][Corporative Network]

原文：http://www.cnblogs.com/qhy285571052/p/61c1f535569220dfb7b920319b6b8935.html