POJ 2785 4 Values whose Sum is 0二分入门

4 Values whose Sum is 0

题目

本题链接
Time Limit:15000MS Memory Limit:228000KB

• Description

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

• Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 2 28 ) that belong respectively to A, B, C and D .

• Output

For each input file, your program has to write the number quadruplets whose sum is zero.

• Sample Input

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45

• Sample Output

5

• Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

分析

显而易见四组数据一一枚举肯定会超时，故本题应采用二分法，将四组数据分别合并成两组数据的和（两个4000*4000数组），进行排序，去重，可二分查找到结果并输出，另外本题也可采用hash做法。

代码

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;int a[4005];int b[4005];int c[4005];int d[4005];int tmp1[16000005];int tmp2[16000005];int re1[16000005][2];int re2[16000005][2];int bs(int key, int num){    int lo=0,hi=num;    int mi;    while(lo<=hi)    {        mi=((hi-lo)>>1)+lo;        if(re1[mi][0]==key) return mi;        else if(re1[mi][0]<key) lo=mi+1;        else hi=mi-1;    }    return -1;}int main(){    int n;    int cnt1 = 0, cnt2 = 0;    int sum = 0;    scanf("%d",&n);    for(int i=0; i<n; i++)        scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);    for(int i=0; i<n; i++)        for(int j=0; j<n; j++){            tmp1[i*n+j] = a[i] + b[j];            tmp2[i*n+j] = c[i] + d[j];        }    sort(tmp1,tmp1+n*n);    sort(tmp2,tmp2+n*n);    re1[0][0] = tmp1[0];    re2[0][0] = tmp2[0];    for(int i=1; i<n*n; i++)    {        if(tmp1[i-1] != tmp1[i])        {            cnt1++;            re1[cnt1][0] = tmp1[i];        }        else if(tmp1[i-1] == tmp1[i])        {            re1[cnt1][1]++;        }        if(tmp2[i-1] != tmp2[i])        {            cnt2++;            re2[cnt2][0] = tmp2[i];        }        else if(tmp2[i-1] == tmp2[i])        {            re2[cnt2][1]++;        }    }    for(int i=0; i<cnt2+1; i++)    {        int t = bs(-re2[i][0],cnt1);        if(t+1)            sum += (re2[i][1]+1) * (re1[t][1]+1);    }    printf("%d\n", sum);    return 0;}