标签:stc nbsp ini class min alt i++ display clu
题意:求一条路径,使得这条边连接1到n,求边权值的最大值与最小值的差
题解:最小生成树,对边权排序,可以枚举边的最大和最小的值,判断能否使得1和n连通
#include <bits/stdc++.h> #define ll long long #define maxn 1010 using namespace std; struct edge{ ll from,to,weight; }; int cmp(edge a,edge b){ return a.weight<b.weight; } vector<edge>edges; int fa[maxn]; int find(int x){ return x==fa[x]?x:(fa[x] = find(fa[x])); } void init(int n){ for(int i=1;i<=n;i++) fa[i] = i; } int main(){ ll n,m,i,j,a,b,c, ma = -1e17,mi = 1e17, ans = 1e18; cin>>n>>m; for(i=0;i<m;i++){ cin>>a>>b>>c; edges.push_back((edge){a,b,c}); } sort(edges.begin(), edges.end(), cmp); for(i=0;i<m;i++){ init(n); mi = 1e17; ma = -1e18; for(j=i;j<m;j++){ edge e = edges[j]; mi = min(e.weight, mi); ma = max(e.weight, ma); int fau = find(e.from); int fav = find(e.to); if(fau != fav) fa[fau] = fav; if(find(n) == find(1)) break; } if(j<m) ans = min(ans, ma-mi); } cout<<ans<<endl; return 0; }
标签:stc nbsp ini class min alt i++ display clu
原文地址:http://www.cnblogs.com/Noevon/p/7162632.html