UVA 10457 Magic Car——最小瓶颈路

来源：互联网发布：越南语发音翻译软件编辑：程序博客网时间：2024/05/23 00:06

其实就是瞎搞，二重循环，第一层枚举最小值，第二层从最小值开始跑MST，当起点和终点在一个集合中时停止，此时求得的边权就是最大值，然后更新结果就好了

PS：第一次刷到vjudge时间第一

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 205;const int maxm = 1005;const int INF = 0x3f3f3f3f;int n, m, scost, ecost, q, s, e, ans;struct Edge {    int from, to, cost;    bool operator < (const Edge &temp) const {        return cost < temp.cost;    }}edge[maxm];int fa[maxn], ran[maxn];void init(int x) {    for (int i = 0; i <= x; i++) fa[i] = i, ran[i] = 0;}int query(int x) { return fa[x] == x ? x : fa[x] = query(fa[x]); }inline void unite(int x, int y) {    x = query(x), y = query(y);    if (x == y) return;    if (ran[x] < ran[y]) fa[x] = y;    else {        fa[y] = x;        if (ran[x] == ran[y]) ran[x]++;    }}int main() {    while (~scanf("%d %d", &n, &m)) {        for (int i = 1; i <= m; i++) scanf("%d %d %d", &edge[i].from, &edge[i].to, &edge[i].cost);        sort(edge + 1, edge + 1 + m);        scanf("%d %d %d", &scost, &ecost, &q);        while (q--) {            ans = INF;            scanf("%d %d", &s, &e);            for (int i = 1; i <= m; i++) {                init(n);                for (int j = i; j <= m; j++) {                    if (edge[j].cost - edge[i].cost + scost + ecost >= ans) break;                    unite(edge[j].from, edge[j].to);                    if (query(s) == query(e)) { ans = min(ans, edge[j].cost - edge[i].cost + scost + ecost); break; }                }            }            printf("%d\n", ans);        }    }    return 0;}

阅读全文

1 0