苗条的生成树（uva 1395）

题目链接 https://vjudge.net/problem/UVA-1395

题意

给出一个n个结点(n<=100)的图，求苗条度尽量小的生成树

思路

把边权从小到大排序，对于一个连续子区间[L,R]，如果能使这n个点联通，那就一定存在一个苗条度不超过W[R]-W[L]的生成树，二重循环枚举即可

#include<bits/stdc++.h>
using namespace std;

const int maxn = 105;
const int maxm = 5050;

int n, m;
int par[maxn];
struct Edge {
int from, to, dist;
}edges[maxm];

bool cmp(Edge x, Edge y) { return x.dist < y.dist; }
int find(int x) { return par[x] == x ? x : par[x] = find(par[x]); }

int main() {
while (scanf("%d%d", &n, &m) == 2) {
if (!n && !m) break;
for (int i = 0; i < m; ++i) scanf("%d%d%d", &edges[i].from, &edges[i].to, &edges[i].dist);
sort(edges, edges + m, cmp);
int ans = -1;
for (int i = 0; i < m; ++i) {
for (int x = 0; x <= n; ++x) par[x] = x;
int num = 0;
for (int j = i; j < m; ++j) {
Edge e = edges[j];
int x = find(e.from);
int y = find(e.to);
if (x != y) {
par[x] = y;
++num;
if (num == n - 1) {
if (-1 == ans) ans = edges[j].dist - edges[i].dist;
else ans = min(ans, edges[j].dist - edges[i].dist);
}
}
}
}
printf("%d\n", ans);
}
return 0;
}