POJ 2785 4 Values whose Sum is 0（排序＋二分搜索）

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 228 ) 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

大致题意：给你Ａ，Ｂ，Ｃ，Ｄ四个数组，在Ａ，Ｂ，Ｃ，Ｄ中各找一个数，使其和为０，求出种数（另外同一数组中相同数值的看成不同数）；

解题思路：这一题大家相信首先想到的是否可以通过枚举得出答案？如果正常枚举的话看一下ｎ的大小，高达４０００，时间复杂度为Ｏ（ｎ＾４）；接着我们来看如何降低时间复杂度！首先我们有ａ＋ｂ＋ｃ＋ｄ＝０可得，只要ａ，ｂ，ｃ确定的话ｄ＝－（ａ＋ｂ＋ｃ），就是说ｄ随之确定。我们只要查找

－（ａ＋ｂ＋ｃ）是否存在于Ｄ中即可，可以先对Ｄ排序，再二分查找实现。先这样优化一下时间复杂度降为Ｏ（ｎ＊ｎ＊ｎ＊ｌｏｇ（ｎ））；这样时间复杂度还是不行的。还要接着优化。最终我们该把Ａ，Ｂ合为一组，Ｃ，Ｄ合为一组，Ａ，Ｂ的组合ｎ＊ｎ种，Ｃ，Ｄ的组合有ｎ＊ｎ种，并存储ｃ＋ｄ，再排序。当确定ａ＋ｂ的和时（有ｎ＊ｎ种）接着从ｃ＋ｄ中二分寻找－（ａ＋ｂ）的值即可，另外可能存在多个值等于－（ａ＋ｂ），这时要稍做处理。

ＡＣ代码：

# include <cstdio>
# include <algorithm>
using namespace std;
int a[4010], b[4010], c[4010], d[4010];
int c_add_d[16000010];
int main(){
	int n, i, j, k, ans, des, left, right, temp, mid, l;
	while(scanf("%d", &n)!=EOF){
		if(n==0){
			printf("0\n");
			continue;
		}
		for(i=1; i<=n; i++){
			scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
		}
		k=1;
		for(i=1; i<=n; i++){
			for(j=1; j<=n; j++){
				c_add_d[k++]=c[i]+d[j];
			}
		}
		sort(c_add_d+1, c_add_d+k);
		ans=0;
		for(i=1; i<=n; i++){
			for(j=1; j<=n; j++){
				left=1; right=k-1; des=-(a[i]+b[j]);
				while(left<=right){
					mid=(left+right)/2;
					if(c_add_d[mid]==des){
						ans++;
						for(l=mid+1; l<=k-1; l++){
							if(c_add_d[l]==des){
								ans++;
							}
							else{
								break;
							}
						}
						for(l=mid-1; l>=1; l--){
							if(c_add_d[l]==des){
								ans++;
							}
							else{
								break;
							}
						}
						break;
					}
					else if(c_add_d[mid]>des){
						right=mid-1;
					}
					else{
						left=mid+1;
					}
				}
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}