UVA 1152 --4 Values whose Sum is 0（枚举--中途相遇法）

4 Values whose Sum is 0

Time Limit:9000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

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 × B × C × D are such that a + b + c + d = 0. In the following, we

assume that all lists have the same size n.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases

following, each of them as described below. This line is followed by a blank line, and there is also a

blank line between two consecutive inputs.

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 228) that belong

respectively to A, B, C and D.

Output

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

The outputs of two consecutive cases will be separated by a blank line.

Sample Input

1

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

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).

题意：大意是有四个集合A,B,C,D从每个中选择一个数，a,b,c,d,使得a+b+c+d = 0，问有多少不同的式子满足条件.

分析：如果数据量小， 直接四重循环即可解决问题，但是数据量太大！所以，考虑所有a+b的和，用一个数组保存所有的值，枚举  -(c+d），有多少和 a+b对应的。

用map会超时！

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 4000+5;
int A[maxn],B[maxn],C[maxn],D[maxn];
int a[maxn*maxn];

int main(){
int T;
scanf("%d",&T);
int flag = 0;
while(T--){
if(flag++)printf("\n");
int n;
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++)
a[i*n+j] = A[i] + B[j];
}

sort(a,a+n*n);
int cnt = 0;

for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
int tmp = C[i] + D[j];
cnt+=upper_bound(a,a+n*n,-tmp) - lower_bound(a,a+n*n,-tmp);
}
}

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

}
return 0;
}
STL中的函数：

1、lower_bound(a,a+n,value)   //返回大于等于value值的第一个位置。

2、upper_bound(a,a+n,value)  //返回大于value值的最后一个值。