poj 2785 二分问题 4 Values whose Sum is 0

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http://poj.org/problem?id=278 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 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). 从每一列里各取一个数,求有多少组四个数的和加起来为零。 先分别求出前两列后两列的所有的和,然后对后两列的和进行排序,对后两列进行二分查找看有多少种情况即可! 枚举会超时的 #include <stdio.h> #include <iostream> #include <string.h> #include <algorithm> using namespace std; int a[4002][4],sum1[16000002],sum2[16000002]; int main() { int n,mid; while(~scanf("%d",&n)) { for(int i=0; i<n; i++) { scanf("%d%d%d%d",&a[i][0],&a[i][1], &a[i][2],&a[i][3]); } int k=0; int m=0; for(int i=0; i<n; i++) for(int j=0; j<n; j++) { sum1[k++]=a[i][0]+a[j][1]; sum2[m++]=a[i][2]+a[j][3]; } sort(sum2,sum2+k); int cnt=0; for(int i=0; i<k; i++) { int left=0; int right=k-1; while(left<=right) { mid=(left+right)/2; if(sum1[i]+sum2[mid]==0) { cnt++; for(int j=mid+1;j<k;j++) { if(sum1[i]+sum2[j]!=0) break; else cnt++; } for(int j=mid-1;j>=0;j--) { if(sum1[i]+sum2[j]!=0) break; else cnt++; } break; } if(sum1[i]+sum2[mid]<0) left=mid+1; else right=mid-1; } } printf("%d\n",cnt); } return 0; }

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