poj 2443 bitset 或 状态压缩

题目链接 http://poj.org/problem?id=2443

Set Operation

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 3235 Accepted: 1317

Description

You are given N sets, the i-th set (represent by S(i)) have C(i) element (Here "set" isn't entirely the same as the "set" defined in mathematics, and a set may contain two same element). Every element in a set is represented by a positive number from 1 to 10000. Now there are some queries need to answer. A query is to determine whether two given elements i and j belong to at least one set at the same time. In another word, you should determine if there exist a number k (1 <= k <= N) such that element i belongs to S(k) and element j also belong to S(k).

Input

First line of input contains an integer N (1 <= N <= 1000), which represents the amount of sets. Then follow N lines. Each starts with a number C(i) (1 <= C(i) <= 10000), and then C(i) numbers, which are separated with a space, follow to give the element in the set (these C(i) numbers needn't be different from each other). The N + 2 line contains a number Q (1 <= Q <= 200000), representing the number of queries. Then follow Q lines. Each contains a pair of number i and j (1 <= i, j <= 10000, and i may equal to j), which describe the elements need to be answer.

Output

For each query, in a single line, if there exist such a number k, print "Yes"; otherwise print "No".

Sample Input

33 1 2 33 1 2 51 1041 31 53 51 10

Sample Output

YesYesNoNo

法1：使用bitset 集合

#include <cstdio>#include <iostream>#include <bitset>using namespace std;bitset<1005> bit[10005];

bitset<1005> tmp;int main(){    int n, m, q, num;    while(~scanf("%d", &n))    {        for(int i=0; i<1001; ++i)            bit[i].reset();//reset全部置0, set全部置1        for(int i=0; i<n; ++i)        {            scanf("%d", &m);            while(m--)            {                scanf("%d", &num);                bit[num][i]=1;            }        }        scanf("%d", &q);        while(q--)        {            int u, v;            scanf("%d %d", &u, &v);            tmp = bit[u]&bit[v];            if(tmp.count()!=0)                puts("Yes");            else                puts("No");        }    }    return 0;}

法2：状态压缩  

由于输入数据范围不超过int， 可以用32位数组存下。

用c++交， G++超时。

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <map>#include <set>#include <queue>#include <vector>#define mem(a) memset(a, 0, sizeof(a))using namespace std;typedef pair<int, int> Pii;typedef long long LL;const int MAXN = 1005;const int inf = 0x3f3f3f3f;const int Mod = 100000000;struct node{    int num[400];    void add(int n)    {        int x=n/31, y=n%31;        num[x]|=1<<y;    }    bool judge(int n)    {        int x=n/31, y=n%31;        return num[x]&(1<<y);    }    void init()    {        memset(num, 0, sizeof(num));    }}a[MAXN];inline int read(){    int f=1, x=0;    char ch = getchar();    while(ch<'0'||ch>'9')    {        if(ch=='-') f=-1;        ch=getchar();    }    while(ch>='0'&&ch<='9')    {        x=x*10+ch-'0';        ch=getchar();    }    return f*x;}int main(){    int n, m, q, k, flag;    while(~scanf("%d", &n))    {        for(int i=1;i<=n;++i)            a[i].init();        for(int i=1;i<=n;++i)        {            scanf("%d", &m);            while(m--)            {                scanf("%d", &k);                a[i].add(k);            }        }        scanf("%d", &q);        int u, v;        while(q--)        {            scanf("%d %d", &u, &v);            flag=0;            for(int i=1;i<=n;++i)            {                if(a[i].judge(u)&&a[i].judge(v))                {                    flag=1;                    break;                }            }            if(!flag)                printf("No\n");            else                printf("Yes\n");        }    }    return 0;}