UVA 10457 - Magic Car(最小瓶颈路)

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UVA 10457 - Magic Car

题目链接

题意：m条路，每条路上必须维持速度v，现在有一辆车，启动能量和结束能量为a, b，途中消耗能量为经过路径最大速度减去最小速度，现在每次循环给定起点终点，问最小能量花费

思路：最小瓶颈路，利用kruskal去搞

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 205;const int M = 1005;int n, m, parent[N];struct Edge {int u, v, val;bool operator < (const Edge& c) const {return val < c.val;}void read() {scanf("%d%d%d", &u, &v, &val);}} E[M];int find(int x) {return x == parent[x] ? x : parent[x] = find(parent[x]);}int solve(int s, int e) {int ans = 1000000000;for (int i = 0; i < m; i++) {for (int j = 1; j <= n; j++) parent[j] = j;for (int j = i; j < m; j++) {int pu = find(E[j].u);int pv = find(E[j].v);if (pu != pv) parent[pu] = pv;if (find(s) == find(e)) {ans = min(ans, E[j].val - E[i].val);break;}}}return ans;}int main() {while (~scanf("%d%d", &n, &m)) {for (int i = 0; i < m; i++)E[i].read();sort(E, E + m);int a, b, q, u, v;scanf("%d%d", &a, &b);scanf("%d", &q);while (q--) {scanf("%d%d", &u, &v);printf("%d\n", a + b + solve(u, v));}}return 0;}

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