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题目大意:给你n个点,m条边,q个询问,每条边有一个val,每次询问也询问一个val,定义:这样条件的两个点(u,v),使得u->v的的价值就是所有的通路中的的最长的边最短。问满足这样的点对有几个。
思路:我们先将询问和边全部都按照val排序,然后我们知道,并查集是可以用来划分集合的,所以我们就用并查集来维护每一个集合就行了。
//看看会不会爆int!数组会不会少了一维! //取物问题一定要小心先手胜利的条件 #include <bits/stdc++.h> using namespace std; #define LL long long #define ALL(a) a.begin(), a.end() #define pb push_back #define mk make_pair #define fi first #define se second const int maxn = 1e5 + 5; const int maxm = 5e5 + 5; int n, m, Q; pair<int, pair<int, int> > p[maxm];///边 pair<int, int> q[maxn];///询问 int par[maxn]; LL sum[maxn], ans[maxn]; int pfind(int x){ if (x == par[x]) return x; return par[x] = pfind(par[x]); } int main(){ while (scanf("%d%d%d", &n, &m, &Q) == 3){ for (int i = 1; i <= m; i++){ int u, v, val; scanf("%d%d%d", &u, &v, &val); p[i] = mk(val, mk(u, v)); } sort(p + 1, p + 1 + m); memset(sum, 0, sizeof(sum)); for (int i = 1; i <= n; i++) par[i] = i, sum[i] = 1; for (int i = 1; i <= Q; i++){ int val; scanf("%d", &val); q[i] = mk(val, i); } sort(q + 1, q + 1 + Q); int j = 1; LL cnt = 0; for (int i = 1; i <= m && j <= Q; ){ while (i <= m && p[i].first <= q[j].first){ int pa = pfind(p[i].second.first); int pb = pfind(p[i].second.second); i++; if (pa == pb) continue; cnt += sum[pa] * sum[pb]; ///printf("sum[%d] = %I64d sum[%d] = %I64d\n", pa, sum[pa], pb, sum[pb]); par[pa] = pb; sum[pb] += sum[pa]; } while (j <= Q && q[j].first < p[i].first){ ans[q[j].second] = cnt; j++; } } while (j <= Q){ ans[q[j].second] = cnt; j++; } for (int i = 1; i <= Q; i++){ printf("%I64d\n", ans[i]); } } return 0; }
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原文地址:http://www.cnblogs.com/heimao5027/p/5873488.html