【Codeforces 1108 F. MST Unification】 最小生成树

题意 给你一个带权无向联通图 他原来的最小生成树边权和为 k 给一条边加 1 是一个操作 问你最少操作多少次使得 这个新连通图的最小生成树 边权和不为k 

做法 我们对边排序

长度一样的里面 如果这两个点已经在一个集合了 你就不用加入最小生成树里面

如果这两个点不在一个集合 说明他有可能加入最小生成树 因为可能加入等值的另一条边也能生成最小生成树 所以你假设给所有等值的边都加上1 然后放入最小生成树的那些边你再撤回 不让他们加 1 这样所得的就是答案了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
int cnt = 0;
const int MAX_N = 200025;
const int MAX_M = 200025;
int fa[MAX_N];
int Find(int x)
{
    if(x==fa[x]) return x;
    return fa[x] = Find(fa[x]);
}
struct edge
{
    int u,v,len;
    edge(){}
    edge(int uu,int vv,int lenn)
    {
        u = uu;
        v = vv;
        len = lenn;
    }
    bool operator < (const edge other) const
    {
        return len < other.len;
    }
}e[MAX_M<<1];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,m,a,b,w,ans = 0;
    scanf("%d%d",&n,&m);
    for(int i = 1;i<=m;++i)
    {
        scanf("%d%d%d",&a,&b,&w);
        e[++cnt] = edge(a,b,w);
    }
    sort(e+1,e+1+cnt);
    for(int i = 1;i<=n;++i) fa[i] = i;
    for(int i = 1,j = 1;i<=cnt;i = j)
    {
        while(j<=cnt&&e[j].len==e[i].len) j++;
        int cnt = j - i;
        for(int t = i;t<j;++t)
        {
            int x = Find(e[t].u),y = Find(e[t].v);
            if(x==y) --cnt;
        }
        for(int t = i;t<j;++t)
        {
            int x = Find(e[t].u),y = Find(e[t].v);
            if(x==y) continue;
            fa[x] = y;
            --cnt;
        }
        ans += cnt;
    }
    printf("%d\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}