【UVa1152】4 Values whose Sum is 0 模拟
2017-12-17 12:33
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原题:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3593
题意:
给出四个集合abcd,求有多少组解。n<=4000。
直接四层循环会超时,所以将等式转换成a[i]+b[i]=-(c[i]+d[i]),预处理一下a[i]+b[i]就可以把复杂度降到O(n^2*logn)。(用upper_bound-lower_bound可以求出相等的个数)
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define lim 4010
using namespace std;
int n,T;
int a[lim],b[lim],c[lim],d[lim],res[lim*lim];
int main()
{
scanf("%d",&T);
while(T--)
{
int tot=0,ans=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++) res[tot++]=a[i]+b[j];//先预处理a[i]+b[i]
sort(res,res+tot);
for (int i=0;i<n;i++)
for (int j=0;j<n;j++) ans+=upper_bound(res,res+tot,-(c[i]+d[j]))-lower_bound(res,res+tot,-(c[i]+d[j]));//这样就求出了相等的个数
printf("%d\n",ans);
if (T) printf("\n");//不加会PE QwQ
}
return 0;
}
题意:
给出四个集合abcd,求有多少组解。n<=4000。
直接四层循环会超时,所以将等式转换成a[i]+b[i]=-(c[i]+d[i]),预处理一下a[i]+b[i]就可以把复杂度降到O(n^2*logn)。(用upper_bound-lower_bound可以求出相等的个数)
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define lim 4010
using namespace std;
int n,T;
int a[lim],b[lim],c[lim],d[lim],res[lim*lim];
int main()
{
scanf("%d",&T);
while(T--)
{
int tot=0,ans=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++) res[tot++]=a[i]+b[j];//先预处理a[i]+b[i]
sort(res,res+tot);
for (int i=0;i<n;i++)
for (int j=0;j<n;j++) ans+=upper_bound(res,res+tot,-(c[i]+d[j]))-lower_bound(res,res+tot,-(c[i]+d[j]));//这样就求出了相等的个数
printf("%d\n",ans);
if (T) printf("\n");//不加会PE QwQ
}
return 0;
}
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