CF 735D

题意:

国际象棋规则,看黑方是否被将军

题解:

八个方向上把最近的点都记下来看看能不能凑合,有一个能凑合就可以输出YES,不然GG

#include<iostream>
#include<cmath>
using namespace std;

const long long INF=1e10+1;

struct S{
    char ch;
    long long x,y;
    S(){}
    S(long long x,long long y){
        this->x=x;
        this->y=y;
    }
}s[501000];

long long nearest[8];
int bit[8];

int main(){
    long long n;
    cin>>n;
    long long x,y;
    cin>>x>>y;
    for(int i=0;i<n;i++){
        cin>>s[i].ch>>s[i].x>>s[i].y;
    }
    for(int i=0;i<8;i++) bit[i]=-1;
    for(int i=0;i<8;i++) nearest[i]=INF;
    for(int i=0;i<n;i++){
        long long disx=abs(s[i].x-x);
        long long disy=abs(s[i].y-y);
        if(s[i].x<x&&s[i].y==y&&disx<nearest[0]){
            nearest[0]=disx;
            bit[0]=i;
        }
        if(s[i].x>x&&s[i].y==y&&disx<nearest[1]){
            nearest[1]=disx;
            bit[1]=i;
        }
        if(s[i].x==x&&s[i].y<y&&disy<nearest[2]){
            nearest[2]=disy;
            bit[2]=i;
        }
        if(s[i].x==x&&s[i].y>y&&disy<nearest[3]){
            nearest[3]=disy;
            bit[3]=i;
        }
        if(disx!=disy) continue;
        if(s[i].x<x&&s[i].y<y&&disx<nearest[4]){
            nearest[4]=disx;
            bit[4]=i;
        }
        if(s[i].x<x&&s[i].y>y&&disx<nearest[5]){
            nearest[5]=disx;
            bit[5]=i;
        }
        if(s[i].x>x&&s[i].y<y&&disx<nearest[6]){
            nearest[6]=disx;
            bit[6]=i;
        }
        if(s[i].x>x&&s[i].y>y&&disx<nearest[7]){
            nearest[7]=disx;
            bit[7]=i;
        }
    }
    for(int i=0;i<8;i++){
        int x=bit[i];
        if(x==-1) continue;
        if(i<4){
            if(s[x].ch=='Q'||s[x].ch=='R'){
                cout<<"YES"<<endl;
                return 0;
            }
        }
        else{
            if(s[x].ch=='Q'||s[x].ch=='B'){
                cout<<"YES"<<endl;
                return 0;
            }
        }
    }
    cout<<"NO"<<endl;
}

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转载自blog.csdn.net/mengwuyaaa/article/details/79967232