hihocoder的hiho一下,第255周的题目
原题
传送门:https://hihocoder.com/contest/hiho255/problem/1
时间限制:10000ms
单点时限:1000ms
内存限制:256MB
描述
There are N queens in an infinite chessboard. We say two queens may attack each other if they are in the same vertical line, horizontal line or diagonal line even if there are other queens sitting between them.
Now given the positions of the queens, find out how many pairs may attack each other?
输入
The first line contains an integer N.Then N lines follow. Each line contains 2 integers Ri and Ci indicating there is a queen in the Ri-th row and Ci-th column.
No two queens share the same position.
For 80% of the data, 1 <= N <= 1000For 100% of the data, 1 <= N <= 100000, 0 <= Ri, Ci <= 1000000000
输出
One integer, the number of pairs may attack each other.样例输入
5 1 1 2 2 3 3 1 3 3 1
样例输出
10
我的解答
这题感觉很简单,每次读入一个皇后的时候,处理一下,横纵斜坐标,她会和横,纵、斜(x+y相同, 或者x-y相同)上原有的皇后都进行attack。代码如下:
#include <iostream>
#include <memory.h>
int main(int argc, char** argv) {
int RSUM[1000000001];
int CSUM[1000000001];
int XIE11[1000000001]; // r-c>=0斜对角
int XIE12[1000000001]; // r-c<0斜对角
int XIE2[2000000002]; // r+c斜对角
memset(RSUM, 0, sizeof(RSUM));
memset(CSUM, 0, sizeof(CSUM));
memset(XIE11, 0, sizeof(XIE11));
memset(XIE12, 0, sizeof(XIE12));
memset(XIE2, 0, sizeof(XIE2));
int n, r, c;
double ans = 0;
scanf("%d", &n);
while(n--) {
scanf("%d %d", &r, &c);
ans = ans + (double)RSUM[r];
RSUM[r]++;
ans = ans + (double)CSUM[c];
CSUM[c]++;
if (r-c >= 0) {
ans = ans + (double)XIE11[r-c];
XIE11[r-c]++;
}
else {
ans = ans + (double)XIE12[c-r];
XIE12[c-r]++;
}
ans = ans + (double)XIE2[r+c];
XIE2[r+c]++;
}
printf("%lf\n", ans);
return 0;
}
但是,不行! 数组开的太大了,RE了,换成malloc也是不行的。
别人的解答
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<map>
using namespace std;
#define ll long long
map<int,int>m1;
map<int,int>m2;
map<int,int>m3;
map<int,int>m4;
int main()
{
int n,a,b,ans=0;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d%d",&a,&b);
ans+=m1[a];
m1[a]++;
ans+=m2[b];
m2[b]++;
ans+=m3[a+b];
m3[a+b]++;
ans+=m4[a-b];
m4[a-b]++;
}
printf("%d\n",ans);
return 0;
}
map用的真好,受教了。话说map没有相应的key的话,是会直接创建一个pair并调用默认的构造函数??