HDU 3047 带权并查集

冲突的条件是：用一个dist数组来保存节点到根的距离，这个距离在路径压缩的时候更新一下，dist[x]+=dist[parent[x]]，然后 在合并时，令r1=Find(u),r2=Find(v),于是合并时就有parent[r2]=r1,dist[r2]=dist[u]+w-dist[v]。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<string>

using namespace std;
typedef long long ll;
const int maxn = 50005;

int n,m;
int par[maxn];
int dis[maxn];

void init(){
memset(par,-1,sizeof(par));
memset(dis,0,sizeof(dis));
}
int find(int x){
int t = x;
int tmp = 0;
bool isRoot = true;
while(par[t] >= 0){
isRoot = false;
tmp += dis[t];
t = par[t];
}
if(isRoot) return x;
else{
dis[x] = tmp;
return par[x] = t;
}
}
bool unite(int x,int y,int k){
int fax = find(x);
//cout<<x<<" "<<fax<<endl;
int fay = find(y);
//cout<<y<<" "<<fay<<endl;
if(fax == fay && dis[y] != dis[x] + k) return true;
if(fax == fay) return false;
par[fay] = fax;
dis[fay] = dis[x] + k - dis[y];
return false;
}
bool same(int x,int y){
return find(x) == find(y);
}

int main(){
while(~scanf("%d%d",&n,&m)){
init();
int a,b,c,ans = 0;
while(m--){
scanf("%d%d%d",&a,&b,&c);
//printf("!!!!!!\n");
if(unite(a,b,c)){
ans++;
//cout<<a<<" "<<b<<" "<<c<<endl;
}
}
printf("%d\n",ans);
}

return 0;
}