http://poj.org/problem?id=1182 (題目鏈接)
題意:中文題
Solution
帶權並查集。
神犇博客,秒懂
%3運用的很巧妙。
代碼:
// poj1182
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf 2147483640
#define Pi 3.1415926535898
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
int fa[1000010],r[1000010],n,k;
int find(int x) {
if (x!=fa[x]) {
int fx=find(fa[x]);
r[x]=(r[x]+r[fa[x]])%3;
fa[x]=fx;
}
return fa[x];
}
bool Union(int x,int y,int type) {
int fx,fy;
fx=find(x);fy=find(y);
if (fx==fy) {
if ((r[y]-r[x]+3)%3!=type) return 1;
else return 0;
}
fa[fy]=fx;
r[fy]=(r[x]-r[y]+type+3)%3;
return 0;
}
int main() {
scanf("%d%d",&n,&k);
for (int i=1;i<=n;i++) fa[i]=i;
memset(r,0,sizeof(r));
int sum=0;
while (k--) {
int d,x,y;
scanf("%d%d%d",&d,&x,&y);
if (x>n || y>n || (x==y && d==2)) sum++;
else if (Union(x,y,d-1)) sum++;
}
printf("%d",sum);
return 0;
}