POJ_2785_4 values whose sum is 0_折半枚举
2014-11-06 19:17
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最近AC率好低啊。。。
题意:
给四个长度为n的数组,问有多少种选择方法,使得从四个数组中各选出一个数字,求和等于0.(数值相同的不同元素也算不同选择方法)
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.
用折半枚举,预处理出后两个数组的和,枚举前两个数组的和,二分搜索。
注意手写二分搜索可能存在找不到符合条件的元素的情况,此时l会等于n-1,应当判断a[l]是否符合情况,不符合就加一,被坑了一次。
代码:
题意:
给四个长度为n的数组,问有多少种选择方法,使得从四个数组中各选出一个数字,求和等于0.(数值相同的不同元素也算不同选择方法)
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.
用折半枚举,预处理出后两个数组的和,枚举前两个数组的和,二分搜索。
注意手写二分搜索可能存在找不到符合条件的元素的情况,此时l会等于n-1,应当判断a[l]是否符合情况,不符合就加一,被坑了一次。
代码:
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> using namespace std; #define mxn 4010 int n,cnt; int a[4][mxn]; int sum[2][mxn*mxn]; int upperbound(int tgt){ int l=0,r=cnt,m; while(l+1<r){ m=(l+r)/2; if(sum[1][m]>tgt) r=m; else l=m; } return sum[1][l]>tgt ? l : l+1; } int lowerbound(int tgt){ int l=0,r=cnt,m; while(l+1<r){ m=(l+r)/2; if(sum[1][m]>=tgt) r=m; else l=m; } return sum[1][l]>=tgt ? l : l+1; } int main(){ scanf("%d",&n); for(int i=0;i<n;++i) for(int j=0;j<4;++j) scanf("%d",&a[j][i]); cnt=0; for(int i=0;i<n;++i) for(int j=0;j<n;++j){ sum[0][cnt]=a[0][i]+a[1][j]; sum[1][cnt++]=a[2][i]+a[3][j]; } sort(sum[0],sum[0]+cnt); sort(sum[1],sum[1]+cnt); int ans=0; for(int i=0;i<cnt;++i) ans+=upperbound(-sum[0][i])-lowerbound(-sum[0][i]); printf("%d\n",ans); return 0; }
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