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 localfreopen("data.txt","r",stdin); #endifwork();return 0; }
