| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 108628 | Accepted: 32960 |
Description
Input
Output
Sample Input
100 7 1 101 1 2 1 2 2 2 3 2 3 3 1 1 3 2 3 1 1 5 5
Sample Output
3
题目中文不解释
虽然我wa了三四发，T了一发（卡scanf），但是感觉这题想明白了还是好写，纸上画画就会了，注意d[]数组在存数的时候别忘记(d[]+3)%3,因为我要保证d只能是三个数,1,2,0;
1代表fa[i]吃i
2反过来
0代表二者同种族
#include <iostream> #include <cmath> #include <cstdio> #include <cstring> #include <string> #include <map> #include <iomanip> #include <algorithm> #include <queue> #include <stack> #include <set> #include <vector> //const int maxn = 1e5+5; #define ll long long ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} #define MAX INT_MAX #define FOR(i,a,b) for( int i = a;i <= b;++i) #define bug cout<<"--------------"<<endl using namespace std; int fa[51000],d[51000]; int cnt,n,m; int Find(int x) { //bug; if(x == fa[x]) return x; int root = Find(fa[x]); d[x] = (d[x] + d[fa[x]]+3)%3; return fa[x] = root; } int main() { ios::sync_with_stdio(false); scanf("%d%d",&n,&m); FOR(i,1,n) fa[i] = i; FOR(i,1,m) { int k,x,y; scanf("%d%d%d",&k,&x,&y); if(x > n || y > n) {cnt++;continue;} if(k==2 && x==y){cnt++;continue;} int fx = Find(x); int fy = Find(y); if(fx == fy) { int temp = d[x] - d[y]; if(k == 1 && temp != 0) { cnt++; continue; } else if(k == 2 ) { if((d[x]-d[y]+3)%3 !=2 ) cnt++; } } else { if(k == 1) { fa[fy] = fx; d[fy] = (d[x] - d[y]+3)%3; } else if(k == 2) { fa[fy] = fx; d[fy] = (d[x] - d[y] + 1+3)%3; } } } printf("%d\n",cnt); }
原文：https://www.cnblogs.com/jrfr/p/11407408.html