2015百度之星 IP聚合

IP聚合

 

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

Problem Description

当今世界，网络已经无处不在了，小度熊由于犯了错误，当上了度度公司的网络管理员，他手上有大量的 IP列表，小度熊想知道在某个固定的子网掩码下，有多少个网络地址。网络地址等于子网掩码与 IP 地址按位进行与运算后的结果，例如：

子网掩码：A.B.C.D

IP 地址：a.b.c.d

网络地址：(A&a).(B&b).(C&c).(D&d)

Input

第一行包含一个整数T，（1≤T≤50）代表测试数据的组数，

接下来T组测试数据。每组测试数据包含若干行，

第一行两个正整数N（1≤N≤1000,1≤M≤50）,M。接下来N行，每行一个字符串，代表一个 IP 地址，

再接下来M行，每行一个字符串代表子网掩码。IP 地址和子网掩码均采用 A.B.C.D的形式，其中A、B、C、D均为非负整数，且小于等于255。

Output

对于每组测试数据，输出两行：

第一行输出: "Case #i:" 。i代表第i组测试数据。

第二行输出测试数据的结果，对于每组数据中的每一个子网掩码，输出在此子网掩码下的网络地址的数量。

Sample Input

2

5 2

192.168.1.0

192.168.1.101

192.168.2.5

192.168.2.7

202.14.27.235

255.255.255.0

255.255.0.0

4 2

127.127.0.1

10.134.52.0

127.0.10.1

10.134.0.2

235.235.0.0

1.57.16.0

Sample Output

Case #1:

3

2

Case #2:

3

4

Problem's Link:   http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=584&pid=1003

---

 

Mean: 

略 

analyse:

贪心

Time complexity: O(n)

 

Source code: 

 

/*

* this code is made by crazyacking

* Verdict: Accepted

* Submission Date: 2015-05-25-14.59

* Time: 0MS

* Memory: 137KB

*/

#include <queue>

#include <cstdio>

#include <set>

#include <string>

#include <stack>

#include <cmath>

#include <climits>

#include <map>

#include <cstdlib>

#include <iostream>

#include <vector>

#include <algorithm>

#include <cstring>

#define  LL long long

#define  ULL unsigned long long

using namespace std;

struct IP

{

    int a, b, c, d;

};

int i, j, k, l, m, n, o, p, q, x, y, z, aa, bb, cc, dd;

struct IP a[1010], b, c[50010], d;

char ch[100];







int dfs()

{

    int i;

    for ( i = 1; i <= q; i++ )

        if ( ( c[i].a == d.a ) && ( c[i].b == d.b ) && ( c[i].c == d.c ) && ( c[i].d == d.d ) ) { return 0; }

    return 1;

}







int main()

{

    scanf( "%d", &n );

    for ( l = 1; l <= n; l++ )

    {

        memset( a, 0, sizeof( a ) );

        memset( c, 0, sizeof( c ) );

        q = 0;

        scanf( "%d%d", &x, &y );

        for ( j = 1; j <= x; j++ )

        {

            scanf( "%s", &ch );

            z = 0; p = 0;

            while ( ( ch[z] > 47 ) && ( ch[z] <= 48 + 9 ) ) {p = p * 10 + ch[z] - 48; z++;}

            a[j].a = p;

            z = z + 1; p = 0;

            while ( ( ch[z] > 47 ) && ( ch[z] <= 48 + 9 ) ) {p = p * 10 + ch[z] - 48; z++;}

            a[j].b = p;

            z = z + 1; p = 0;

            while ( ( ch[z] > 47 ) && ( ch[z] <= 48 + 9 ) ) {p = p * 10 + ch[z] - 48; z++;}

            a[j].c = p;

            z = z + 1; p = 0;

            while ( ( ch[z] > 47 ) && ( ch[z] <= 48 + 9 ) ) {p = p * 10 + ch[z] - 48; z++;}

            a[j].d = p;

        }

        printf( "Case #%d:\n", l );

        for ( j = 1; j <= y; j++ )

        {

            q = 0;

            memset( c, 0, sizeof( c ) );

            scanf( "%s", &ch );

            z = 0; p = 0;

            while ( ( ch[z] > 47 ) && ( ch[z] <= 48 + 9 ) ) {p = p * 10 + ch[z] - 48; z++;}

            b.a = p;

            z = z + 1; p = 0;

            while ( ( ch[z] > 47 ) && ( ch[z] <= 48 + 9 ) ) {p = p * 10 + ch[z] - 48; z++;}

            b.b = p;

            z = z + 1; p = 0;

            while ( ( ch[z] > 47 ) && ( ch[z] <= 48 + 9 ) ) {p = p * 10 + ch[z] - 48; z++;}

            b.c = p;

            z = z + 1; p = 0;

            while ( ( ch[z] > 47 ) && ( ch[z] <= 48 + 9 ) ) {p = p * 10 + ch[z] - 48; z++;}

            b.d = p;

            for ( k = 1; k <= x; k++ )

            {

                aa = ( a[k].a ) & ( b.a );

                bb = ( a[k].b ) & ( b.b );

                cc = ( a[k].c ) & ( b.c );

                dd = ( a[k].d ) & ( b.d );

                d.a = aa; d.b = bb; d.c = cc; d.d = dd;

                if ( q == 0 ) {q++; c[q] = d; continue;}

                if ( dfs() ) {q++; c[q] = d;}

            }

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

        }

    }

    return 0;

}

View Code