UVALive 4035 - Undetectable Tour（并查集）

Posted 2020-08-20 TianTengtt

题意：给定一个 N * N（3 <= N <= 10000）的矩形区域，左下角为（0，0），右上角为（N，N），现在要从左下角走到右上角，但是有 k（k <= 100）个监视器，每个监视器的监视范围都是统一的，现给定监视范围可能出现的种类与概率，求能够逃出去的概率。（计算距离时均用曼哈顿距离）

1、若两个监视器的距离 <= 两个监视器的监视范围之和，则这两个监视器间的路就不可走；

2、若监视器距离墙的距离 <= 监视器的监视范围，则监视器与墙之间就不可走；

综上，对于每一个监视范围，可以借助并查集判断是否 "左边界和上边界" 与 "右边界和下边界" 之间不可走，若可走则证明在此监视范围下可以逃出去。

若将监视范围排序，并二分找临界答案，时间将更短。

代码如下：

 

  1 #pragma comment(linker, "/STACK:102400000, 102400000")
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<cctype>
  5 #include<cstdlib>
  6 #include<cmath>
  7 #include<iostream>
  8 #include<sstream>
  9 #include<iterator>
 10 #include<algorithm>
 11 #include<string>
 12 #include<vector>
 13 #include<set>
 14 #include<map>
 15 #include<deque>
 16 #include<queue>
 17 #include<stack>
 18 #include<list>
 19 #define fin freopen("in.txt", "r", stdin)
 20 #define fout freopen("out.txt", "w", stdout)
 21 #define pr(x) cout << #x << " : " << x << "   "
 22 #define prln(x) cout << #x << " : " << x << endl
 23 #define Min(a, b) ((a < b) ? a : b)
 24 #define Max(a, b) ((a < b) ? b : a)
 25 typedef long long ll;
 26 typedef unsigned long long llu;
 27 const int INT_INF = 0x3f3f3f3f;
 28 const int INT_M_INF = 0x7f7f7f7f;
 29 const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
 30 const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
 31 const double pi = acos(-1.0);
 32 const double EPS = 1e-8;
 33 const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
 34 const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
 35 const ll MOD = 1e9 + 7;
 36 using namespace std;
 37 
 38 #define NDEBUG
 39 #include<cassert>
 40 const int MAXN = 100 + 10;
 41 const int MAXT = 10000 + 10;
 42 
 43 struct Detector{
 44     int x, y;
 45     Detector(int xx = 0.0, int yy = 0.0) : x(xx), y(yy) {}
 46 };
 47 
 48 struct Query{
 49     int radius;
 50     double pro;
 51     bool operator < (const Query &rhs) const{
 52         return radius < rhs.radius;
 53     }
 54 }q[MAXN];
 55 
 56 int T, m, fa[MAXN * 3], n;
 57 vector<Detector> vec;
 58 
 59 int Find(int x) {
 60     return fa[x] = (fa[x] == x) ? x : Find(fa[x]);
 61 }
 62 
 63 inline int get_dist(int x1, int y1, int x2, int y2) {
 64     return abs(x1 - x2) + abs(y1 - y2);
 65 }
 66 
 67 inline void merge_(int x, int y) {
 68     int one = Find(x), two = Find(y);
 69     if(one == two)  return;
 70     if(one < two)  fa[two] = one;
 71     else  fa[one] = two;
 72 }
 73 
 74 bool judge(int dist) {
 75     for(int i = 0; i < MAXN * 3; ++i)  fa[i] = i;
 76     int len = (int)vec.size() - 1;
 77     for(int i = 1; i <= len; ++i)
 78         for(int j = i + 1; j <= len; ++j) {
 79             int tmp = get_dist(vec[i].x, vec[i].y, vec[j].x, vec[j].y);
 80             if(tmp <= dist * 2)  merge_(i, j);
 81         }
 82     for(int i = 1; i <= len; ++i) {
 83         if(vec[i].x <= dist || n - vec[i].y <= dist)  merge_(0, i);
 84         if(vec[i].y <= dist || n - vec[i].x <= dist)  merge_(i, len + 1);
 85     }
 86     return Find(0) != Find(len + 1);
 87 }
 88 
 89 int solve() {
 90     int l = 0, r = m - 1;
 91     while(l < r) {
 92         int mid = l + (r - l + 1) / 2;
 93         if(judge(q[mid].radius))  l = mid;
 94         else  r = mid - 1;
 95     }
 96     return judge(q[l].radius) ? l : -1;
 97 }
 98 
 99 int main(){
100 //    fin;
101     scanf("%d", &T);
102     while(T--) {
103         vec.clear();
104         vec.push_back(Detector());
105         scanf("%d%d", &n, &m);
106         for(int i = 0; i < m; ++i)  scanf("%d%lf", &q[i].radius, &q[i].pro);
107         sort(q, q + m);
108         int u, v;
109         while(scanf("%d", &u) == 1 && u != -1) {
110             scanf("%d", &v);
111             vec.push_back(Detector(u, v));
112         }
113         int len = solve();
114         double ans = 0.0;
115         for(int i = 0; i <= len; ++i)  ans += q[i].pro;
116         printf("%g\n", ans);
117     }
118     return 0;
119 }