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http://www.lydsy.com/JudgeOnline/problem.php?id=1196
题目大意:n个城市,m-1条路,每条路有一级公路和二级公路之分,你要造n-1条路,一级公路至少要造k条,求出所造路的最大所需的val的最小值.
思路:首先我们一定要明确这个不是一题求所有花费的最小值的问题。然后我们只要二分答案就可以了。最后注意一下条件的拜访即可。
//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 = 10000 + 5; const int inf = 0x3f3f3f3f; struct Edge{ int u, v, val1, val2; Edge(int u = 0, int v = 0, int v1 = 0, int v2 = 0): u(u), v(v), val1(v1), val2(v2){} bool operator < (const Edge &a) const{ if (val1 != a.val1) return val1 < a.val1; return val2 < a.val2; } }e[maxn * 2]; int n, k, m; int par[maxn]; int pfind(int x){ if (par[x] == x) return x; return par[x] = pfind(par[x]); } bool judge(int midval){ for (int i = 1; i <= n; i++) par[i] = i; int cnt1 = 0, cnt = 0; for (int i = 1; i <= m - 1; i++){ Edge a = e[i]; int pu = pfind(a.u), pv = pfind(a.v); if (pu == pv) continue; if (a.val1 <= midval) cnt1++, cnt++, par[pu] = pv; else if (a.val2 <= midval) cnt++, par[pu] = pv; } if (cnt == n-1 && cnt1 >= k) return true; return false; } int main(){ scanf("%d%d%d", &n, &k, &m); int lb = 0, rb = 30000; for (int i = 1; i <= m - 1; i++){ int u, v, v1, v2; scanf("%d%d%d%d", &u, &v, &v1, &v2); e[i] = Edge(u, v, v1, v2); } sort(e + 1, e + m); while (lb < rb){ int mid = lb + (rb - lb) / 2; if (judge(mid)){ rb = mid; } else { lb = mid + 1; } } printf("%d\n", lb); return 0; } /* 3 1 3 1 2 11 4 2 3 10 2 */
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原文地址:http://www.cnblogs.com/heimao5027/p/5857102.html