Connections in Galaxy War(反向并查集)

最新推荐文章于 2022-01-09 21:53:50 发布

原创最新推荐文章于 2022-01-09 21:53:50 发布 · 545 阅读

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本文介绍了一种使用反向并查集解决特定问题的方法。通过解析代码，展示了如何处理星系之间的连接，并针对一系列查询提供解答。文章包括了完整的C++代码实现，涉及结构体定义、查找和合并操作。

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题目链接上一种法:http://blog.youkuaiyun.com/qq_37753409/article/details/77922448.
提交后发现Runtime error. 查阅资料后,发现要是用反向并查集,在这里用反向并查集实现.
题目链接:https://vjudge.net/problem/ZOJ-3261
代码如下:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
using namespace std;
const int STAR_SIZE = 10010;
const int INQUIRY_SIZE = 50010;
struct Star{
    int ID;
    int POWER;
    void setData(int id, int power){
        ID = id;
        POWER = power;
    }
};
struct Inquiry{
    int a, b;
    int flag;
    int ans;
    void setData(int f, int aa, int bb){
        a = aa;
        b = bb;
        flag = f;
    }
};
Star star[STAR_SIZE];
Inquiry inquiry[INQUIRY_SIZE];
int MAX[STAR_SIZE];
set< pair<int, int> > T;
void init(){
    T.clear();
    for(int i = 0; i < STAR_SIZE; ++i){
        star[i].setData(i, -1);
    }
    for(int i = 0; i < INQUIRY_SIZE; ++i){
        inquiry[i].setData(-1, 0, 0);
        inquiry[i].ans = -1;
    }
    memset(MAX, -1, sizeof(MAX));
} 
int find(int p){
    if(star[p].ID != p){
        int d = star[p].ID;
        star[p].ID = find(star[p].ID);
        MAX[p] = max(MAX[p], MAX[d]);
    }
    return star[p].ID;
}
void Union(int a, int b){
    int x = find(a);
    int y = find(b);
    if(x == y) return;
    if(MAX[x] < MAX[y] || (MAX[x] == MAX[y] && x > y)){
        star[x].ID = y;
    }else{
        star[y].ID = x;
    }   
}
int main(){
    int n, m, q;
    int k = 0;
    while(~scanf("%d", &n)){
        if(k++)puts("");
        init();
        char op[10];
        int a, b;
        for(int i = 0; i < n; ++i){
            scanf("%d", &star[i].POWER);
        }
        scanf("%d", &m);
        for(int i = 0; i < m; ++i){
            scanf("%d%d", &a, &b);
            if(a > b) swap(a, b);           
            T.insert(make_pair(a, b));
        }
        scanf("%d", &q);
        for(int i = 0; i < q; ++i){
            scanf("%s", op);
            if(op[0] == 'q'){
                scanf("%d", &a);
                inquiry[i].setData(0, a, 0);
            }else if(op[0] == 'd'){
                scanf("%d%d", &a, &b);
                if(a > b) swap(a, b);
                T.erase(make_pair(a, b));
                inquiry[i].setData(1, a, b);
            }
        }

        for(int i = 0; i < n; ++i){
            MAX[i] = star[i].POWER;
        }

        for(set< pair<int, int> >::iterator it = T.begin(); it != T.end(); it++){
            a = (*it).first;
            b = (*it).second;
            Union(a, b);
        }
        for(int i = q - 1; i >= 0; i--){
            if(inquiry[i].flag == 0){
                if(MAX[find(inquiry[i].a)] > star[inquiry[i].a].POWER){
                    inquiry[i].ans = star[inquiry[i].a].ID;
                }else{
                    inquiry[i].ans = -1;
                }
            }else if(inquiry[i].flag == 1){
                Union(inquiry[i].a, inquiry[i].b);
            }
        }
        for(int i = 0;i < q; i++)
        {
            if(inquiry[i].flag == 0)
            {
                printf("%d\n",inquiry[i].ans);
            }               
        }
    }
    return 0;
}