POJ 1379 模拟退火

Posted 2021-01-10 bigtom

Run Away

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10634		Accepted: 3177

Description

One of the traps we will encounter in the Pyramid is located in the Large Room. A lot of small holes are drilled into the floor. They look completely harmless at the first sight. But when activated, they start to throw out very hot java, uh ... pardon, lava. Unfortunately, all known paths to the Center Room (where the Sarcophagus is) contain a trigger that activates the trap. The ACM were not able to avoid that. But they have carefully monitored the positions of all the holes. So it is important to find the place in the Large Room that has the maximal distance from all the holes. This place is the safest in the entire room and the archaeologist has to hide there.

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing three integers X, Y, M separated by space. The numbers satisfy conditions: 1 <= X,Y <=10000, 1 <= M <= 1000. The numbers X and Yindicate the dimensions of the Large Room which has a rectangular shape. The number M stands for the number of holes. Then exactly M lines follow, each containing two integer numbers Ui and Vi (0 <= Ui <= X, 0 <= Vi <= Y) indicating the coordinates of one hole. There may be several holes at the same position.

Output

Print exactly one line for each test case. The line should contain the sentence "The safest point is (P, Q)." where P and Qare the coordinates of the point in the room that has the maximum distance from the nearest hole, rounded to the nearest number with exactly one digit after the decimal point (0.05 rounds up to 0.1).

Sample Input

3
1000 50 1
10 10
100 100 4
10 10
10 90
90 10
90 90
3000 3000 4
1200 85
63 2500
2700 2650 
2990 100

Sample Output

The safest point is (1000.0, 50.0).
The safest point is (50.0, 50.0).
The safest point is (1433.0, 1669.8).

Source

Central Europe 1999

思路：

同POJ2420， 更改了下每次步长与X，Y值相关就过了，要不然死活过不了样例。

还有衰减系数取0.99过不了样例，取0.999可以。

过了样例就A了。

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define pli pair<ll, int>
 29 #define pil pair<int, ll>
 30 #define clr(a, x) memset(a, x, sizeof(a))
 31 
 32 const double pi = acos(-1.0);
 33 const int INF = 0x3f3f3f3f;
 34 const int MOD = 1e9 + 7;
 35 const double EPS = 1e-9;
 36 
 37 /*
 38 #include <ext/pb_ds/assoc_container.hpp>
 39 #include <ext/pb_ds/tree_policy.hpp>
 40 using namespace __gnu_pbds;
 41 #define TREE tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update>
 42 TREE T;
 43 */
 44 
 45 int T, n;
 46 struct Point {
 47     double x, y;
 48     Point () {}
 49     Point (double x, double y) : x(x), y(y) {}
 50     double dis(Point p) {
 51         return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));
 52     }
 53 } p[1005];
 54 inline double get_p(double delta, double T) {
 55     return exp(delta / T);
 56 }
 57 double cal(Point pp) {
 58     double res = 1e18;
 59     for (int i = 1; i <= n; ++i) {
 60         double tres = pp.dis(p[i]);
 61         res = min(res, tres);
 62     }
 63     return res;
 64 }
 65 const int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
 66 double X, Y;
 67 Point ans_p;
 68 double tuihuo() {
 69     double x = rand() % (int)(X + 1), y = rand() % (int)(Y + 1);
 70     double res = cal(Point(x, y)), ans = res;
 71     ans_p = Point(x, y);
 72     double T = 10000 / min(X, Y), phi = 0.999;
 73     for (int rep = 1; rep <= 100000; ++rep) {
 74         int r = rand() % 4;
 75         int deltax = dir[r][0], deltay = dir[r][1];
 76         double tox = x + deltax * T * X;
 77         double toy = y + deltay * T * Y;
 78         if (0 <= tox && tox <= X && 0 <= toy && toy <= Y) {
 79             double tres = cal(Point(tox, toy));
 80             if (tres > res) {
 81                 x = tox, y = toy;
 82                 res = tres;
 83                 if (res > ans) {
 84                     ans = res;
 85                     ans_p = Point(x, y);
 86                 }
 87             } else {
 88                 double delta = tres - res;
 89                 double prob = get_p(delta, T);
 90                 prob = max(0.1, prob);
 91                 if (rand() < prob * RAND_MAX) {
 92                     x = tox, y = toy;
 93                     res = tres;
 94                 }
 95             }
 96         }
 97         T *= phi;
 98     }
 99     return ans;
100 }
101 int main() {
102     srand(19980305);
103     scanf("%d", &T);
104     while (T--) {
105         scanf("%lf%lf%d", &X, &Y, &n);
106         for (int i = 1; i <= n; ++i) {
107             scanf("%lf%lf", &p[i].x, &p[i].y);
108         }
109         tuihuo();
110         printf("The safest point is (%.1f, %.1f).
", ans_p.x, ans_p.y);
111     }
112     return 0;
113 }

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