2016 Multi-University Training Contest 3 1011【鸽巢原理】
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题解:
坐标(0,m)的话,闭区间,可能一共有多少曼哈顿距离?
2m
但是给一个n,可能存在n(n+1)/2个曼哈顿距离
所以可以用抽屉原理了
当n比抽屉的数量大,直接输出yes
不用计算
那。。。NO呢
那就暴力
n^2
这个n肯定不会太大
队友code……..
#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>#include <math.h>#include <queue>#include <stack>using namespace std;#define INF 0x3f3f3f#define pi acos(-1.0)#define MAX 200010#define mod 9973#define ll long longint n,m;int dist[MAX];struct node{ int x,y;}d[MAX];bool cmp(node a,node b){ if(a.x==b.x) return a.x<b.x; return a.y<b.y;}int main(){ int t,i,j,k; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); memset(dist,0,sizeof(dist)); for(i=0;i<n;i++) scanf("%d%d",&d[i].x,&d[i].y); sort(d,d+n,cmp); ll sum=2*m; if(n>2*m) { printf("YES\n"); continue; } ll cont=0; int flag=0; for(i=0;i<n;i++) { for(j=i+1;j<n;j++) { cont++; int a=abs(d[i].x-d[j].x)+abs(d[i].y-d[j].y); dist[a]++; if(dist[a]>1||cont>MAX) { flag=1; break; } } if(flag) break; } if(flag) printf("YES\n"); else printf("NO\n"); } return 0;}
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