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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;
}

UVA 10457 - Magic Car(最小瓶颈路)

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原文地址：http://blog.csdn.net/accelerator_/article/details/38897735