并查集 poj1611&poj2492

poj1611 简单题

代码中id记录父节点，sz记录子树规模。一个集合为一棵树。

#include <iostream>

#include <cstdio>

using namespace std;



int id[300005];

int sz[300005];



void add(int a, int b)

{

    int i, j;

    for (i = a; i != id[i]; i = id[i]);

    for (j = b; j != id[b]; j = id[j]);

    if (i == j) return;

    if (sz[i] < sz[j]) {

        id[i] = j;

        sz[j] += sz[i];

    } else {

        id[j] = i;

        sz[i] += sz[j];

    }

}



int main()

{

    int n, m, cnt, root, temp;

    while (scanf("%d%d", &n, &m) != EOF) {

        if (!n) break;

        for (int i = 0; i < n; ++i) {

            id[i] = i;

            sz[i] = 1;

        }

        for (int i = 0; i < m; ++i) {

            scanf("%d", &cnt);

            scanf("%d", &root);

            for (int j = 1; j < cnt; ++j) {

                scanf("%d", &temp);

                add(root, temp);

            }

        }

        int ans = 0;

        int rt;

        for (rt = 0; rt != id[rt]; rt = id[rt]);

        for (int i = 0; i < n; ++i) {

            int j;

            for (j = i; j != id[j]; j = id[j]);

            if (j == rt) ++ans;

        }

        printf("%d\n", ans);

    }

    return 0;

}

　　

poj 2492

题目也是醉了，看半天没看懂= =#

输入每对a b表示a和b是夫妻，问有没有同性恋= =

把每一次a b放入同一集合，并用rel记录每个节点和它父节点的相对关系。这样同一集合的任意两点间关系就确定了

/**********************************************

Memory: 8500 KB		Time: 125 MS

Language: G++		Result: Accepted

***********************************************/

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;



int id[2010], sz[2010];

int a[1000005], b[1000005];

int rel[2010];  //和父节点的性别是否一致，一致为0，否则为1



int Scan() {    //输入外挂

    int res = 0, flag = 0;

    char ch;

    if((ch = getchar()) == '-') flag = 1;

    else if(ch >= '0' && ch <= '9') res = ch - '0';

    while((ch = getchar()) >= '0' && ch <= '9')

        res = res * 10 + (ch - '0');

    return flag ? -res : res;

}



void add(int a, int b)

{

    int i, j;

    int rel_a = 0, rel_b = 0;

    for (i = a; i != id[i]; i = id[i])

        rel_a = (rel_a + rel[i]) % 2;　　//a和根节点的关系

    for (j = b; j != id[j]; j = id[j])

        rel_b = (rel_b + rel[j]) % 2;　　//b和根节点的关系

    if (i == j) return ;

    if (sz[i] <= sz[j]) {   //i->j

        sz[j] += sz[i];

        id[i] = j;

        rel[i] = (rel_a == rel_b) ? 1 : 0;



    } else { //j->i

        sz[i] += sz[j];

        id[j] = i;

        rel[j] = (rel_a == rel_b) ? 1 : 0;

    }

}



int is_gay(int a, int b)

{

    int i, j;

    int rel_a = 0, rel_b = 0;

    for (i = a; i != id[i]; i = id[i])

        rel_a = (rel_a + rel[i]) % 2;

    for (j = b; j != id[j]; j = id[j])

        rel_b = (rel_b + rel[j]) % 2;

    if (i == j && rel_a == rel_b)

        return 1;

    return 0;

}





int main()

{

    int t, m, n;

    t = Scan();

    for (int k = 1; k <= t; ++k) {

        n = Scan();

        m = Scan();

        for (int i = 1; i <= n; ++i) {

            id[i] = i;

            sz[i] = 1;

            rel[i] = 0;

        }

        for (int i = 0; i < m; ++i) {

            a[i] = Scan();

            b[i] = Scan();

            add(a[i], b[i]);

        }

        int i;

        for (i = 0; i < m; ++i) {

            if (is_gay(a[i], b[i]))

                break;

        }

        printf("Scenario #%d:\n%s\n\n", k, i == m ? "No suspicious bugs found!" : "Suspicious bugs found!");

    }

    return 0;

}

　　

总结经验教训：以后多敲两行也不能复制粘贴= =太坑。。。