标签:span using bit else problem color scan min bool
题目链接: https://www.lydsy.com/JudgeOnline/problem.php?id=1050
思路:
先将每条边的权值排个序优先小的,然后从小到大枚举每一条边,将其存到并查集里,如果得到的比值比之前的小,那么判断下s与t能否连通,如果连通就替换就好了
实现代码:
#include<bits/stdc++.h> using namespace std; const int M = 1e6+10; int f[M],vis[M],a[M]; int n,m; int Find(int x){ if(x==f[x])return x; return f[x]=Find(f[x]); } void mix(int x,int y){ int fx = Find(x); int fy = Find(y); if(fx != fy) f[fx] = fy; } bool cmp(int a,int b){ return a > b; } struct node{ int x,y,v; bool operator < (const node &cmp) const{ return v < cmp.v; } }e[M]; int main() { int s,t; scanf("%d%d",&n,&m); for(int i = 1;i <= m;i ++) scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].v); sort(e+1,e+1+m); scanf("%d%d",&s,&t); int minn = 1,maxx = 30000; for(int i = 1;i <= m;i ++){ for(int j = 1;j <= n;j ++) f[j] = j; for(int j = i;j <= m;j ++){ mix(e[j].x,e[j].y); if(e[j].v*minn > e[i].v*maxx) break; if(Find(s) == Find(t)){ int k = __gcd(e[j].v,e[i].v); minn = e[i].v/k; maxx = e[j].v/k; // cout<<k<<" "<<minn<<" "<<maxx<<endl; break; } } } if(maxx == 30000&&minn == 1) printf("IMPOSSIBLE\n"); else if(minn == 1) printf("%d\n",maxx); else cout<<maxx<<"/"<<minn<<endl; return 0; }
标签:span using bit else problem color scan min bool
原文地址:https://www.cnblogs.com/kls123/p/9346166.html