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。。。