ACM Contest and Blackout

ACM Contest and Blackout - UVA 10600 - 次小生成树

题意

  n个点，m条边，求最小生成树的值和次小生成树的值。

思路

  无脑写次小生成树就OK，因为比较简单我就尝试着用Kruskal写了一下，本以为耗时会很长的，结果是0ms，感觉可能是数据水的原因。

代码

////  main.cpp//  L////  Created by LucienShui on 2017/7/2.//  Copyright © 2017年 LucienShui. All rights reserved.//#include <iostream>#include <algorithm>#include <set>#include <string>#include <vector>#include <queue>#include <map>#include <stack>#include <iomanip>#include <cstdio>#include <cstring>#include <cmath>#include <cctype>#define memset(a,b) memset(a,b,sizeof(a))#define il inline#define ll long long#define ull unsigned long longusing namespace std;#define maxn 107const int INF = 0x3f3f3f3f;int T,n,m,ans,ans2,pre[maxn],fa[maxn];struct node {    int a, b, c;    bool operator < (const node& tmp) const {        return c < tmp.c;    }}a[maxn*maxn];vector<node> used;int find(int x, int pre[]) { return pre[x] == x ? x : pre[x] = find(pre[x], pre); }void solve() {    sort(a, a+m);    int cnt = 1;    ans = 0;    used.clear();    for(int i=1;i<=n;i++) pre[i] = i;    for(int i=0;i<m;i++) {        int x = find(a[i].a, pre), y = find(a[i].b, pre);        if(x != y) {            pre[x] = y;            used.push_back(a[i]);            ++cnt;            ans += a[i].c;        }        if(cnt == n) break;    }    ans2 = INF;    for(int i=0;i<used.size();i++) {        node v = used[i];        int cur = 0, cnt = 1;        for(int j=1;j<=n;j++) fa[j] = j;        for(int j=0;j<m;j++) {            if(v.a == a[j].a && v.b == a[j].b && v.c == a[j].c) continue;            int x = find(a[j].a, fa), y = find(a[j].b, fa);            if(x != y) {                fa[x] = y;                ++cnt;                cur += a[j].c;            }            if(cnt == n) break;        }        if(cnt == n) ans2 = min(ans2, cur);    }}int main() {#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endif    scanf("%d",&T);    while(T--) {        scanf("%d%d",&n,&m);        for(int i=0;i<m;i++) scanf("%d%d%d",&a[i].a,&a[i].b,&a[i].c);        solve();        printf("%d %d\n",ans,ans2);    }    return 0;}