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

[b]Mean:[/b]

略

[b]analyse:[/b]

贪心

[b]Time complexity: O(n)[/b]

[b]Source code: [/b]

/*
* 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