1050: [HAOI2006]旅行comf

1050: [HAOI2006]旅行comf

链接

分析

　　考虑如何统计答案，且复杂度较小，直接搜索是不行的。那么考虑先枚举一条较小的边，然后依次加入比它大的点，一旦ST联通了，那么说明当前比值为 最新加入的边 比 枚举的较小的边，用并查集维护，取最小值。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
 
inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}
 
const int N = 5005;
 
struct Edge{
    int u,v,w;
    bool operator < (const Edge &A) const {
        return w < A.w;
    }
}e[N];
int fa[505];
int gcd(int a,int b) {
    if (b == 0) return a;
    return gcd(b,a%b);
}
int find(int x) {
    if (x == fa[x]) return x;
    return fa[x] = find(fa[x]);
}
 
int main() {
    int n = read(),m = read();
    for (int i=1; i<=m; ++i) {
        e[i].u = read(),e[i].v = read(),e[i].w = read();
    }
    int S = read(),T = read();
    int ans1 = -1,ans2 = -1;
    double ans = 1e9;
    sort(e+1,e+m+1);
     
    for (int i=1; i<=m; ++i) {
        for (int j=1; j<=n; ++j) fa[j] = j;
         
        for (int j=i; j<=m; ++j) {
            int u = find(e[j].u),v = find(e[j].v);
            if (u != v) fa[u] = v;
             
            if (find(S) == find(T)) {
                double t = (1.0 * e[j].w) / (1.0 * e[i].w);
                if (t < ans) {
                    ans = t;ans1 = e[i].w,ans2 = e[j].w;
                }
                break;
            }
        }
        if (ans1 == -1 && ans2 == -1) {
            puts("IMPOSSIBLE");return 0;
        }
    }
    int d = gcd(ans1,ans2);
    ans2 /= d,ans1 /= d;
    if (ans1 == 1) cout << ans2;
    else cout << ans2 << "/" << ans1; //居然写成了 cout << ans2/d << "\" << ans1/d !!!     
    return 0;
}