D - Cat and Mice Kattis - catandmice （状压+二分）

题目链接：

https://cn.vjudge.net/problem/Kattis-catandmice

题目大意：

自己翻译，懒得写

具体思路：

二分枚举最佳速度，然后状压dp判断是否合理。

AC代码：

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 #define ll long long
  4 #define inf 0x3f3f3f3f
  5 #define LL_inf (1ll << 60)
  6 const int maxn = 2e5 + 100;
  7 const int mod = 1e9 + 7;
  8 const double eps = 1e-5;
  9 struct node
 10 {
 11     double x, y, t;
 12     node() {}
 13     node(double t1, double t2, double t3)
 14     {
 15         x = t1;
 16         y = t2;
 17         t = t3;
 18     }
 19 } q[maxn];
 20 double dp[34000][20];
 21 char str[10];
 22 int n;
 23 double dis(double x1, double y1, double x2, double y2)
 24 {
 25     return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
 26 }
 27 int cnt[maxn];
 28 int cal(int t)
 29 {
 30     int ans = 0;
 31     for (int i = 0; i < 16; i++)
 32     {
 33         ans += ((t & (1 << i)) > 0 ? 1 : 0);
 34     }
 35     return ans;
 36 }
 37 void init()
 38 {
 39     int maxstate = (1 << n) - 1;
 40     for (int i = 0; i <= maxstate; i++)
 41     {
 42         cnt[i] = cal(i);
 43     }
 44 }
 45 double qsm(double t1, int t2)
 46 {
 47     double ans = 1.0;
 48     while (t2)
 49     {
 50         if (t2 & 1)
 51             ans = ans * t1;
 52         t2 >>= 1;
 53         t1 = t1 * t1;
 54     }
 55     return ans;
 56 }
 57 bool check(double t, double lv)
 58 {
 59     for(int i=0; i<(1<<n); i++)
 60         for(int j=0; j<n; j++)
 61             dp[i][j]=inf;
 62     int maxstate = (1 << n) - 1;
 63     for (int i = 0; i < n; i++)
 64     {
 65         if (dis(q[i].x, q[i].y, 0.0, 0.0) / t <= q[i].t)
 66             dp[(1 << i)][i] = dis(q[i].x, q[i].y, 0.0, 0.0) / t;
 67     }
 68     for (int i = 1; i <= maxstate; i++)
 69     {
 70         for (int j = 0; j < n; j++)
 71         {
 72             for (int k = 0; k < n; k++)
 73             { // 这里加判断(i&(1<<k))，时间差4倍多，，
 74                 if ((i & (1 << j)) &&((i & (1 <<k))==0) &&(dis(q[j].x, q[j].y, q[k].x, q[k].y) / (t * qsm(lv, cnt[i])) + dp[i][j] <= q[k].t))
 75                 {
 76                     dp[i | (1 << k)][k] =
 77                         min(dp[i | (1 << k)][k],
 78                             dis(q[j].x, q[j].y, q[k].x, q[k].y) / (t * qsm(lv, cnt[i] ))  + dp[i][j]);
 79                 }
 80             }
 81         }
 82     }
 83     for (int i = 0; i < n; i++)
 84     {
 85         if (dp[maxstate][i] <= q[i].t)
 86             return true;
 87     }
 88     return false;
 89 }
 90 int main()
 91 {
 92     scanf("%d", &n);
 93     init();
 94     for (int i = 0; i < n; i++)
 95     {
 96         scanf("%lf %lf %lf", &q[i].x, &q[i].y, &q[i].t);
 97     }
 98     double lv;
 99     cin>>lv;
100     double l = 0, r = 1e9;
101     double ans = 0;
102     while (r - l >=eps)
103     {
104         double mid = (l + r) / 2.0 ;
105         if (check(mid, lv))
106         {
107             ans=mid;
108             r = mid;
109         }
110         else
111             l = mid;
112     }
113     printf("%.3f\n", ans);
114     return 0;
115 }

 

posted @ 2019-06-21 15:34  Let_Life_Stop  阅读(209)  评论(0)  编辑  收藏  举报