POJ 1182 食物链(带权并查集)

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题目链接：POJ 1182 食物链

没学带权并查集时候做过一次这个题，没做出来，然后看了题解被那些看起来很巧妙的通项公式吓到了，放弃了。现在再做这个题，发现这个题是可以不用那些通项公式的，完全可以把几种情况列出来做。

其实把一些情况列出来之后连蒙带猜也是可以得出来通项公式的，并没有我开始想的那么复杂。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAX_N = 50000 + 100;int n, k, p[MAX_N], _rank[MAX_N];int _find(int x){    if(p[x] != x)    {        int root = _find(p[x]);        _rank[x] = (_rank[x] + _rank[p[x]]) % 3;        return p[x] = root;    }    return x;}int main(){    scanf("%d%d", &n, &k);    for(int i = 1; i <= n; i++)        p[i] = i;    int o, a, b, u, v, res = 0;    for(int i = 0; i < k; i++)    {        scanf("%d%d%d", &o, &a, &b);        if((o == 2) && (a == b) || a > n || b > n)        {            res++;            continue;        }        u = _find(a);        v = _find(b);        if(u == v)        {            /*if(o == 1)                if(!(_rank[a] == _rank[b]))                    res++;            if(o == 2)                if(!((_rank[b] == 1 && _rank[a] == 2) || (_rank[b] == 2 && _rank[a] == 0) || (_rank[b] == 0 && _rank[a] == 1)))                    res++;*/             if((_rank[a] - _rank[b] + 3) % 3 != o - 1)                res++;            continue;        }        p[u] = v;        _rank[u] = (_rank[b] - _rank[a] + o - 1 + 3) % 3;    }    printf("%d\n", res);    return 0;}

算是今天才彻底搞懂吧。看了http://blog.csdn.net/niushuai666/article/details/6981689 15/11/17

/**relation[x] 表示rootx->x的值 注意不要搞反x->y=1 x吃yx->y=2 y吃xx->y=0 x与y是同类根据向量得到所求relation的值*/#include <iostream>#include <cstdio>#include <cstring>const int MAX_N = 50000 + 100;using namespace std;struct Ani{    int pa, re;};Ani anis[MAX_N];int _find(int x){    if(x == anis[x].pa)        return x;    int tempf = anis[x].pa;    anis[x].pa = _find(anis[x].pa);    anis[x].re = (anis[tempf].re + anis[x].re) % 3;    return anis[x].pa;}int main(){    int n, k, sum = 0;    scanf("%d%d", &n, &k);    for(int i = 1; i <= n; i++)    {        anis[i].pa = i;        anis[i].re = 0;    }    int d, x, y;    for(int i = 0; i < k; i++)    {        scanf("%d%d%d", &d, &x, &y);        if(x > n || y > n)        {            sum++;            continue;        }        if(d == 1)        {            int xf = _find(x);            int yf = _find(y);            if(xf != yf)            {                anis[xf].pa = yf;                anis[xf].re = (anis[y].re + 3 - anis[x].re) % 3;            }            else            {                int now = (3 - anis[x].re + anis[y].re) % 3;                if(now != d - 1)                    sum++;            }        }        else        {            int xf = _find(x);            int yf = _find(y);            if(xf != yf)            {                anis[xf].pa = yf;                anis[xf].re = (anis[y].re + 3 - (d - 1) + 3 - anis[x].re) % 3;            }            else            {                int now = (3 - anis[x].re + anis[y].re) % 3;                if(now != d - 1)                    sum++;            }        }    }    printf("%d\n", sum);    return 0;}

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