UESTC 1594

題意：中文題

思路：和poj食物鏈的題幾乎一樣，拆點或者帶權並查集做，這種分類不多的比較傾向與拆點做

AC代碼：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define
 
 mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

int pre[N<<1],n,m;
void init(int n){
    for(int i=0; i<=3*n; ++i) pre[i]=i;
}
int finds(int
 
 x){
    return pre[x]=pre[x]==x?x:finds(pre[x]);
}
void unions(int x, int y){
    int fx=finds(x), fy=finds(y);
    pre[fy]=fx;
}
int main(){
    //ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>m;
    init(n);
    int t=0;
    for(int i=1; i<=m; ++i){
        int t,x,y;
        scanf(
 
"%d%d%d",&t,&x,&y);
        if(x>n || y>n || x<1 || y<1 || (t==2 && x==y)){
            printf("%d ",i);
            continue;
        }
        if(t==1){
            if(finds(x)==finds(y+n) || finds(x)==finds(y+2*n)){
                printf("%d ",i);
            }
            else{
                unions(x,y);
                unions(x+n,y+n);
                unions(x+2*n,y+2*n);
            }
        }
        else{
            if(finds(x)==finds(y) || finds(x)==finds(y+2*n)){
                printf("%d ",i);
            }
            else{
                unions(x,y+n);
                unions(x+n,y+2*n);
                unions(x+2*n,y);
            }
        }
    }
    return 0;
}

UESTC 電子科大專題訓練 數據結構 L