1050: [HAOI2006]旅行comf
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 5 inline int read() { 6 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 7 for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; 8 } 9 10 const int N = 5005; 11 12 struct Edge{ 13 int u,v,w; 14 bool operator < (const Edge &A) const { 15 return w < A.w; 16 } 17 }e[N]; 18 int fa[505]; 19 int gcd(int a,int b) { 20 if (b == 0) return a; 21 return gcd(b,a%b); 22 } 23 int find(int x) { 24 if (x == fa[x]) return x; 25 return fa[x] = find(fa[x]); 26 } 27 28 int main() { 29 int n = read(),m = read(); 30 for (int i=1; i<=m; ++i) { 31 e[i].u = read(),e[i].v = read(),e[i].w = read(); 32 } 33 int S = read(),T = read(); 34 int ans1 = -1,ans2 = -1; 35 double ans = 1e9; 36 sort(e+1,e+m+1); 37 38 for (int i=1; i<=m; ++i) { 39 for (int j=1; j<=n; ++j) fa[j] = j; 40 41 for (int j=i; j<=m; ++j) { 42 int u = find(e[j].u),v = find(e[j].v); 43 if (u != v) fa[u] = v; 44 45 if (find(S) == find(T)) { 46 double t = (1.0 * e[j].w) / (1.0 * e[i].w); 47 if (t < ans) { 48 ans = t;ans1 = e[i].w,ans2 = e[j].w; 49 } 50 break; 51 } 52 } 53 if (ans1 == -1 && ans2 == -1) { 54 puts("IMPOSSIBLE");return 0; 55 } 56 } 57 int d = gcd(ans1,ans2); 58 ans2 /= d,ans1 /= d; 59 if (ans1 == 1) cout << ans2; 60 else cout << ans2 << "/" << ans1; //居然写成了 cout << ans2/d << "\" << ans1/d !!! 61 return 0; 62 }