HLG1067QQ Farm【状压dp】

最新推荐文章于 2016-04-18 23:36:00 发布

原创最新推荐文章于 2016-04-18 23:36:00 发布 · 269 阅读

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本文针对QQ农场游戏中的蚊子拍击问题进行算法分析。通过使用状压DP算法解决每秒至少拍击一只蚊子以获取经验值的问题，并考虑了拍击蚊子数量与得分的关系。

QQ Farm

Time Limit: 1000 MS

Memory Limit: 65536 K

Total Submit: 33(11 users)

Total Accepted: 10(8 users)

Rating:

Special Judge: No

Description

Many boys and girls play the game QQ farm, For example, in this picture we can see xnby、 Lance and so on. Sometimes there are some mosquitoes in our farm, we use a rectangular-shaped pad to kill them, and our experience will increase by 3 points after killing each mosquito .But now we find that using a square-shaped pad works better. when using the square-shaped pad, if we kill n mosquito at one time, our experience will increase by n×n points. For example, if we kill 4 mosquitoes at one time, our experience will increase by 16 points. A mosquito was killed when it is covered by the pad, but killing them is not so easy, because them are flying all the time. Now we got the radius of the pad and the location of every mosquito in each second, in each second we could use the pad just once, and we must kill at least one mosquito every second. You are asked to write a program to compute how many points we can get at most.

Input

There are multiple test cases.

In each case, The first line contains two integers: N, T, which indicate the number of mosquitoes and the number of seconds.

The second line contains one Real number R, the length of the pad’s side.

The following T lines, each contains N pairs of Real numbers: (X₁, Y₁), (X₂, Y₂)…(X_T, Y_T), which indicate the locations of those mosquitoes at the corresponding second.

Those real numbers are rounded to two digits after the decimal point.

The sides of the pad must be parallel to x-axis or Y-axis.

( 0 < N ≤ 10, 0 < T ≤ N, -1000000 ≤ R, Xi, Yi ≤ 1000000 )

Output

The number of points we could get at most.

Sample Input

3 2

2.00

1.00 1.00 2.00 2.00 3.00 3.00

0.00 0.00 2.00 2.00 4.00 4.00

3 1

2.00

1.00 1.00 2.00 2.00 3.00 3.00

Sample Output

Author

yiyi4321@Footmen

啊啊啊啊啊啊啊啊这个题卡我两天啊啊啊啊啊啊啊啊啊啊啊

大意：有n个蚊子 t个时期告诉你每个时期每个蚊子的坐标

有一个正方形的苍蝇拍每个时期最少拍一个蚊子 point+拍死蚊子个数的额平方

问最后最多的point

分析：

由于n<= 10所以状压

dp[i][j] 代表i时期状态为j最大的point

dp[i][j] = max(dp[i - 1][k] + cnt * cnt)

然后这时候遇到一个问题就是苍蝇拍该如何放的问题

刚开始的时候我是这么处理的

枚举所有的一对点i，j然后求出以这两个点位边缘的正方形有包含哪几个点

然后wa一次

原因是不可能每一次都拍的最大的点的个数有可能它选择一个少的个数的拍

然后我又想到一个思路就是用一个vector存所有的拍的情况

vector v[i][j]代表 i时期最多拍死j个蚊子有哪些团

这时候就能用10*10来枚举所有的状态

然后又wa无数次

之后跟wht讨论中发现了错误

原来每次枚举拍死x个蚊子的时候它可能拍死的不止x个蚊子

例如1011他拍死的肯能是1111但是不会被发现

处理就是枚举一下所有的点看其它点是不是也被收入之中若没有被收入那么就是合法的删除状态

代码：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <vector>
  5 #include <cmath>
  6 using namespace std;
  7 
  8 const int maxn = 11;
  9 const double INF = 1000000000.0;
 10 int dp[maxn][1 << maxn];
 11 vector<int>v[maxn][maxn];
 12 int ok[maxn];
 13 double x[maxn][maxn], y[maxn][maxn];
 14 int t, n;
 15 double r;
 16 int eps(double x) {
 17     if(fabs(x) < 1e-6) {
 18         return 0;
 19     } 
 20     if(x < 0) return -1;
 21     return 1;
 22 }
 23 
 24 void init() {
 25     for(int i = 1; i < maxn; i++) {
 26         for(int j = 1; j < maxn; j++) {
 27             v[i][j].clear();
 28         }
 29     }
 30     int Max = 1 << n;
 31     double nx, ny, mx, my;
 32     for(int i = 1; i <= t; i++) {
 33         for(int j = 0; j < Max; j++) {
 34             int cnt = 0; nx = ny = INF; mx = my = - INF;
 35             for(int k = 1; k <= n; k++) {
 36                 if((j & ( 1 << ( k - 1 ) ) )) {
 37 //                    printf(" j == %d  k == %d   ans = %d\n", j, k, (j & ( 1 << ( k - 1 ) ) ));
 38                     ok[cnt++] = k;
 39                     nx = min(nx, x[i][k]);
 40                     ny = min(ny, y[i][k]);
 41 
 42                     mx = max(mx, x[i][k]);
 43                     my = max(my, y[i][k]);
 44                 }
 45             }
 46             if(cnt && eps(mx - nx - r) <= 0 && eps(my - ny - r) <= 0) {
 47                 int vis[maxn] = { 0 };
 48                 bool flag = true;
 49                 for(int k = 0; k < cnt; k++) {
 50                     vis[ok[k]] = 1;
 51                 }
 52                 for(int l = 1; l <= n; l++) {
 53                     if(!vis[l] && eps(x[i][l] - nx) >= 0 && eps(x[i][l] - mx) <= 0 && eps(y[i][l] - ny) >= 0 && eps(y[i][l] - my) <= 0) {
 54                         flag = false;
 55                         break;
 56                     }
 57                 }
 58                 if(flag) 
 59                     for(int k = 0; k < cnt; k++) {
 60                         v[i][cnt].push_back(ok[k]);
 61                     }
 62             }
 63         }
 64     }
 65 //    for(int i = 1; i <= t; i++) {
 66 //        for(int j = 1; j <= n; j++) {
 67 //            for(int k = 0; k < v[i][j].size(); k++) {
 68 //                printf("%d %d %d\n", i, j, v[i][j][k]);
 69 //            }
 70 //        }
 71 //    }
 72 //    puts("");
 73 }
 74 
 75 void DP() {
 76     memset(dp, -1, sizeof(dp));
 77     dp[0][0] = 0;
 78     int Max = 1 << n;
 79     for(int i = 1; i <= t; i++) {
 80         for(int j = 0; j < Max; j++) {
 81             if(dp[i - 1][j] != -1) {
 82                 for(int k = 1; k <= n; k++) {
 83                     int vs = v[i][k].size();
 84                     if(vs) {
 85                         int flag = j; int cnt = 0;
 86                         for(int l = 0; l < vs; l++) {
 87                             int id = v[i][k][l];
 88 //                            printf("i == %d  k == %d  ans = %d\n", i, k, v[i][k][l]);
 89                             if((flag & ( 1 << ( id - 1 ) ) ) == 0) {
 90                                 flag |= ( 1 << ( id - 1 ) );
 91                                 cnt ++;
 92                             }
 93                             if((l + 1) % k == 0 && cnt) {
 94                                 dp[i][flag] = max(dp[i][flag], dp[i - 1][j] + cnt * cnt);
 95                                 flag = j;
 96                                 cnt = 0;
 97                             }
 98                         }
 99                     }
100                 }
101             }
102         }
103     }
104 }
105 
106 int ANS() {
107     int ans = 0; int Max = 1 << n;
108     for(int i = 0; i < Max; i++) {
109         ans = max(ans, dp[t][i]);
110     }
111     return ans;
112 }
113 
114 int main() {
115     while(EOF != scanf("%d %d",&n, &t) ) {
116         scanf("%lf",&r);
117         for(int i = 1; i <= t; i++) {
118             for(int j = 1; j <= n; j++) {
119                 scanf("%lf %lf",&x[i][j], &y[i][j]);
120             }
121         }
122         init();
123         DP();
124         printf("%d\n", ANS());
125     }
126     return 0;
127 }

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