Is There A Second Way Left? - UVA 10462 - 次小生成树

Is There A Second Way Left? - UVA 10462 - 次小生成树

题意

思路

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 long

using namespace std;

#define maxn 107

const int INF = 0x3f3f3f3f;

int pre[maxn],used[maxn<<1],result,cnt;

struct Edge {
int x,y,v;
bool operator < (const Edge tmp) const {
return v < tmp.v;
}
}edge[maxn<<1];

void init(int n) {
for(int i=0 ; i<=n ; i++) pre[i] = i;
}

int Find(int x) {//while以及附帶路徑壓縮的版本，遞歸在數據量比較大的時候可能爆
if(x==pre[x]) return x;
int p=x,q;
while(x!=pre[x]) x=pre[x];
while(p!=pre[p])
q=pre[p],pre[p]=x,p=q;
return x;
}

int kruskal(int n, int m) {
int ans = cnt = 0;
init(n);
sort(edge,edge+m);
for(int i=0 ; i<m ; i++) {
int p = Find(edge[i].x), q = Find(edge[i].y);
if(p!=q) {
pre[p] = q;
ans += edge[i].v;
used[cnt++] = i;
}
}
if(cnt+1 == n) return ans;
return INF;
}

int kruskal_(int n, int m, int del) {
int ans = 0, temp = 0;
init(n);
for(int i=0 ; i<m ; i++) {
if(i == del) continue;
int p = Find(edge[i].x), q = Find(edge[i].y);
if(p!=q) {
pre[p] = q;
ans += edge[i].v;
temp++;
}
}
if(temp+1 == n) return ans;
return INF;
}

int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
int t,n,m,cas{};
scanf("%d",&t);
while(t--) {
printf("Case #%d : ",++cas);
scanf("%d%d",&n,&m);
for(int i=0 ; i<m ; i++)
scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].v);
if(kruskal(n,m)==INF) {
printf("No way\n");
continue;
}
result = INF;
for(int i=0 ; i<cnt ; i++)
result = min(result, kruskal_(n,m,used[i]));
if(result == INF)
printf("No second way\n");
else printf("%d\n",result);
}
return 0;
}