D. Anton and Chess 模拟题 + 读题

http://codeforces.com/contest/734/problem/D

一开始的时候看不懂题目，以为象是中国象棋那样走，然后看不懂样例。

原来是走对角线的，长知识了。

所以我们就知道，王有八个方向，所以每个方向选一个来做代表就行了。

那么选谁呢？可以排序，按照他们离王的距离从小到大排，这样就能选出最近的那个（不用被棋挡住）

然后注意下方向的表达，在王的右上角，还要和王在同一对角线才行~不能简单地判x和y的大小关系

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 500000 + 20;
LL xx0, yy0;
struct node {
    char ch;
    LL x, y;
    node() {}
    node(char cc, LL xx, LL yy) : ch(cc), x(xx), y(yy) {}
    bool operator < (const struct node & rhs) const {
        LL dis = (x - xx0) * (x - xx0) + (y - yy0) * (y - yy0);
        LL dis2 = (rhs.x - xx0) * (rhs.x - xx0) + (rhs.y - yy0) * (rhs.y - yy0);
        return dis < dis2;
    }
}arr[maxn];
vector<struct node>pos[66];
void work() {
    IOS;
    int n;
    cin >> n;
    cin >> xx0 >> yy0;
    for (int i = 1; i <= n; ++i) {
        char str[11];
        cin >> str;
        arr[i].ch = str[0];
        cin >> arr[i].x >> arr[i].y;
    }
    sort(arr + 1, arr + 1 + n);
//    for (int i = 1; i <= n; ++i) {
//        cout << arr[i].ch << " " << arr[i].x << " " << arr[i].y << endl;
//    }
    for (int i = 1; i <= n; ++i) {
        int face = 0;
        if (arr[i].x == xx0 && arr[i].y > yy0) face = 1;
        else if (arr[i].x < xx0 && arr[i].y > yy0 && arr[i].x + arr[i].y == xx0 + yy0) face = 2;
        else if (arr[i].y == yy0 && arr[i].x < xx0) face = 3;
        else if (arr[i].x < xx0 && arr[i].y < yy0 && arr[i].x - arr[i].y == xx0 - yy0) face = 4;
        else if (arr[i].x == xx0 && arr[i].y < yy0) face = 5;
        else if (arr[i].x > xx0 && arr[i].y < yy0 && arr[i].x + arr[i].y == xx0 + yy0) face = 6;
        else if (arr[i].y == yy0 && arr[i].x > xx0) face = 7;
        else if (arr[i].x > xx0 && arr[i].y > yy0 && arr[i].x - arr[i].y == xx0 - yy0) face = 8;
        if (pos[face].size() != 0) continue;
        pos[face].push_back(arr[i]);
    }
//    cout << "fff" << endl;
    for (int i = 1; i <= 8; ++i) {
        if ((i == 1 || i == 5) && pos[i].size() && (pos[i][0].ch == 'Q' || pos[i][0].ch == 'R')) {
            printf("YES\n");
            return;
        } else if ((i == 2 || i == 6) && pos[i].size() && (pos[i][0].ch == 'Q' || pos[i][0].ch == 'B')) {
            printf("YES\n");
            return;
        } else if ((i == 3 || i == 7) && pos[i].size() && (pos[i][0].ch == 'Q' || pos[i][0].ch == 'R')) {
            printf("YES\n");
            return;
        } else if ((i == 4 || i == 8) && pos[i].size() && (pos[i][0].ch == 'B' || pos[i][0].ch == 'Q')) {
            printf("YES\n");
            return;
        }
    }
    printf("NO\n");
}

int main() {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    work();
    return 0;
}