Sunday, June 12, 2016

[hihoCoder] #1307 : 穿越禁区 解题报告 - Jonah的专栏 - 博客频道 - CSDN.NET


[hihoCoder] #1307 : 穿越禁区 解题报告 - Jonah的专栏 - 博客频道 - CSDN.NET
作为H国的精英特工,你接到了一项任务,驾驶一辆吉普穿越布满监测雷达的禁区。为了简化题目,我们可以把禁区想象为一个左下角是(0, 0)右上角是( WH )的长方形区域。区域中一共有 N 座雷达,其中第 i 座的坐标是(Xi,Yi ),监测范围是半径为 Ri 的圆形区域。所有在圆内和圆上的运载工具都会被监测到。
你的目标是从左到右穿越禁区。你可以选择线段(0, 0)-(0, H)上任意一点作为起点,线段(W, 0)-(WH)上任意一点作为终点。在禁区内你可以沿任意路线行驶,只要保持始终在禁区内并且没有被雷达监测到。
给出禁区内的雷达部署方案,你需要判断是否存在满足条件的行驶路线。

输入

输入包含多组数据。
第1行是一个整数 T,表示以下有 T 组数据 (1 ≤ T ≤ 10)。
每组数据的第1行:三个整数 WHN (0 ≤ WH ≤ 1000000, 1 ≤ N ≤ 1000)。
每组数据的第2-N+1行:每行三个整数XiYiRi (0 ≤ Xi ≤ W, 0 ≤ Yi ≤ H, 1 ≤ Ri ≤ 1000000)。

输出

对于每组数据输出"YES"或者"NO"表示是否有满足条件的行驶路线。

样例输入
2
10 4 2
5 1 1
5 3 1
10 4 2
5 1 1
6 3 1
样例输出
NO
YES

思路: 这道是并查集的题目, 有好多个雷达, 我们要把能够连在一起的雷达合并起来, 看其总共是否能够覆盖从上到下. 先给每个雷达默认父结点是本身, 然后依次遍历每个雷达, 如果A雷达和B雷达能够接触到, 即他们圆心距离小于两个半径之和, 说明他们在一个集合里面, 就把B的最深父结点设置为A的最深父结点. 这样做完之后, 再依次将每个雷达的父结点设为本身的最深父结点, 就可以知道总共有几个集合, 然后计算每个集合所覆盖的范围是多少即可.
int main() { vector<pair<int ,int> > vec; int T, W, H, N, xi, yi, ri; cin >> T; while(T--) { cin >> W >> H >> N; vector<vector<int> > radars; for(int i = 0; i < N; i++) { cin >> xi >> yi >> ri; radars.push_back(vector<int>{xi, yi, ri}); } int ans = 1; vector<int> hash(N); for(int i = 0; i < N ;i++) hash[i] = i; for(int i =0; i< N; i++) { int x1 = radars[i][0], y1 = radars[i][1], r1 = radars[i][2]; for(int j = i+1; j < N; j++) { int x2 = radars[j][0], y2 = radars[j][1], r2 = radars[j][2]; float dis = sqrt(pow(x1-x2, 2)+ pow(y1-y2, 2)); if(dis <= r1+r2) { int par1 = hash[j], par2 = hash[i]; while(hash[par1] != par1) par1 = hash[par1]; while(hash[par2] != par2) par2 = hash[par2]; hash[par1] = par2; } } } set<int> parent; for(int i =0; i< N; i++) { int par = hash[i]; while(par != hash[par]) par = hash[par]; hash[i] = par; parent.insert(par); } for(int val: parent) { int top = INT_MIN, bot = INT_MAX; for(int i = 0; i < N; i++) { if(hash[i] == val) { top = max(top, radars[i][1] + radars[i][2]); bot = min(bot, radars[i][1] - radars[i][2]); if(top >= H && bot <=0) { ans = 0; i = N; break; } } } if(ans == 0) break; } if(ans) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }

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