HDU 5441 Travel （带权并查集 离线）

题目

HDU 5441 Travel

题意

n个点，m条边，每条边权值val
q次询问，每次有一个值v,将图中所有小于v的边标记出来组成新图，求里面有多少个点对可以相互到达(a -> b 和 b -> a 算两个)

解法

离线并查集，记录每个集合里面点的个数，每次合并两个点数为x, y的集合，对ans的贡献为 \((x+y)\*(x+y-1) - x\*(x-1) - y\*(y-1)\) 整理得\(2xy\)

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 2e4 + 100;
const int M = 1e5 + 100;
const int K = 5e3 + 100;
struct Edge {
    int u, v, val;
    friend bool operator < (const Edge& a, const Edge& b) {
        return a.val < b.val;
    }
} edge[M];
struct Question {
    int val, pos;
    friend bool operator < (const Question& a, const Question& b) {
        return a.val < b.val;
    }
} q[K];
int fa[N], ans[K], num[N];
int find(int u) {
    if(fa[u] == u) {
        return u;
    }
    else {
        return fa[u] = find(fa[u]);
    }
}
int main() {
    int T;
    int n, m, k;
    scanf("%d", &T);
    while(T--) {
        scanf("%d %d %d", &n, &m, &k);
        for(int i = 0; i < m; i++) {
            scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].val);
        }
        for(int i = 0; i < k; i++) {
            scanf("%d", &q[i].val);
            q[i].pos = i;
        }
        sort(edge, edge + m);
        sort(q, q + k);
        for(int i = 0; i <= n; i++) {
            fa[i] = i;
            num[i] = 1;
        }
        int j = 0, sum = 0;
        for(int i = 0; i < k; i++) {
            while(j < m && edge[j].val <= q[i].val) {
                int fu = find(edge[j].u);
                int fv = find(edge[j].v);
                if(fu != fv) {
                    int x = num[fu];
                    int y = num[fv];
                    fa[fv] = fu;
                    // sum -= x * (x - 1);
                    // sum -= y * (y - 1);
                    // sum += (x + y) * (x + y - 1);
                    sum += 2 * x * y;
                    num[fu] += y;
                    num[fv] = 0;
                }
                j++;
            }
            ans[q[i].pos] = sum;
        }
        for(int i = 0; i < k; i++) {
            printf("%d\n", ans[i]);
        }
    }
    return 0;
}

来源

2015 ACM/ICPC Asia Regional Changchun Online

赛后总结

2015 ACM/ICPC Asia Regional Changchun Online