POJ 1182

膜大牛~寫的超好，很詳細，對於偏移量的確定也有講解~http://blog.csdn.net/niushuai666/article/details/6981689

記得用scanf……cin……T了……orz

#include<iostream>
#include<stdio.h>
using namespace std;
const int N = 5e4 + 5;
struct node {
    int pre;
    int relation;
};
node p[N];
int n, d,k, x, y,ans;
void init(int n)
{
    
 
for (int i = 0; i <= n; i++)
    {
        p[i].pre = i;
        p[i].relation = 0;
    }
}
int findx(int x)  //帶勸啥的還是老老實實用遞歸……
{
    if (x == p[x].pre)
     return x;
    int temp = p[x].pre;
    p[x].pre = findx(temp);
    p[x].relation = (p[x].relation + p[temp].relation) % 3;
    return p[x].pre;
 
}
void unionn(int a, int b)
{
    x = findx(a);
    y = findx(b);
    if (x != y)
    {
        p[y].pre = x;
        p[y].relation = (3+ p[a].relation+d-1-p[b].relation) % 3;
    }
    else
    {
        if (d == 1 && p[a].relation != p[b].relation)
            ans++;
        else if (d == 2
 
 && (3 - p[a].relation + p[b].relation) % 3 != d - 1)
            ans++;
    }
}
int main()
{
    scanf("%d%d", &n, &k);
    init(n);
    ans = 0;
    while (k--)
    {
        scanf("%d%d%d", &d, &x,&y);
        if (x > n || y > n) ans++;
        else if (x == 2 && x == y) ans++;
        else unionn(x, y);
    }
    printf("%d\n", ans);
    return 0;
}

POJ 2492

這個要比食物鏈簡單很多，只需要區分♂♀兩類，emmmm其實上一個也可以用下面的方法做

#include<iostream>
#include<stdio.h>
using namespace std;
int n, t,k, a, b,flag;
int pre[4005];
void init(int n)
{
    for (int i=0; i <= 2*n; i++)
        pre[i] = i;
}
int findx(int x)
{
    if (x == pre[x]) return x;
    int r = pre[x];
    pre[x] = findx(r);
    return pre[x];
}
void unionn(int a, int b)
{
    int x = findx(a), y = findx(b);
    if(x!=y)
    pre[y] = x;
}
bool judge(int x, int y)
{
    if (findx(x) == findx(y)) return 0;
    else return 1;
}
int main()
{
    scanf("%d", &t);
    int cnt = 1;
    while (t--)
    {
        scanf("%d%d", &n, &k);
        init(n);
        flag = 0;
        for (int i = 0; i < k; i++)
        {
            scanf("%d%d", &a, &b);
            if (judge(a, b) || judge(a + n, b + n))
            {
                unionn(a, b+n);
                unionn(a + n, b);
            }
            else flag = 1;
        }
                if(cnt!=1)   //不要忘記換行orz
                 puts("");
        if (flag)
        {
            printf("Scenario #%d:\n", cnt++);
            puts("Suspicious bugs found!");
                 }
        else
        {
            printf("Scenario #%d:\n", cnt++);
            puts("No suspicious bugs found!");
        }
    }
    return 0;

}

POJ 1182 食物鏈 種類並查集