本文共 2575 字，大约阅读时间需要 8 分钟。
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 #include #include #include #include #define IOS ios::sync_with_stdio(false)using namespace std;#define inf (0x3f3f3f3f)typedef long long int LL;#include #include #include #include #include #include #include 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 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;}
转载于:https://www.cnblogs.com/liuweimingcprogram/p/6068366.html
