POJ 1191 棋盘分割

题意：中文题，感觉没什么好翻译的。http://poj.org/problem?id=1191

解法：将均方差公式变形，用ave表示平均值，均方差 = (∑xi*xi - 2 * ave * ∑xi + n * ave * ave) / n = (∑xi*xi - n*ave*ave)/n。由于平均数一定，所以只需要使平方之和最小即可。

　　　设s[x1][y1][x2][y2]表示以(x1, y1)为左上角，以(x2, y2)为右下角的矩形里，所有小格的数之和的平方。d[t][x1][y1][x2][y2]表示切t刀，以(x1, y1)为左上角，以(x2, y2)为右下角的矩形所能参生的最小平方和是多少。

tag：dp

 1 /*
 2  * Author:  Plumrain
 3  * Created Time:  2013-11-17 19:22
 4  * File Name: DP-POJ-1191.cpp
 5  */
 6 #include <iostream>
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <cmath>
10 
11 using namespace std;
12 
13 #define CLR(x) memset(x, 0, sizeof(x))
14 const int maxint = 2147483647;
15 
16 int n;
17 double sum;
18 int d[20][10][10][10][10];
19 int a[10][10], ss[10][10][10][10], s[10][10][10][10];
20 
21 void init()
22 {
23     sum = 0.0;
24     for (int i = 1; i <= 8; ++ i)
25         for (int j = 1; j <= 8; ++ j){
26             scanf ("%d", &a[i][j]);
27             sum += a[i][j];
28         }
29 
30     CLR (ss);
31     for (int i = 1; i <= 8; ++ i)
32         for (int j = 1; j <= 8; ++ j)
33             for (int k = i; k <= 8; ++ k)
34                 for (int l = j; l <= 8; ++ l){
35                     if (i == k && j == l) 
36                         ss[i][j][k][l] = a[i][j];
37                     else if (j == l)
38                         ss[i][j][k][l] = ss[i][j][k-1][l] + a[k][l];
39                     else{
40                         ss[i][j][k][l] = ss[i][j][k][l-1];
41                         for (int ii = i; ii <= k; ++ ii)
42                             ss[i][j][k][l] += a[ii][l];
43                     }
44                 }
45 
46     CLR (s);
47     for (int i = 1; i <= 8; ++ i)
48         for (int j = 1; j <= 8; ++ j)
49             for (int k = i; k <= 8; ++ k)
50                 for (int l = j; l <= 8; ++ l)
51                     s[i][j][k][l] = ss[i][j][k][l]*ss[i][j][k][l];
52 }
53 
54 double gao()
55 {
56     for (int i = 1; i <= 8; ++ i)
57         for (int j = 1; j <= 8; ++ j)
58             for (int k = i; k <= 8; ++ k)
59                 for (int l = j; l <= 8; ++ l)
60                     d[1][i][j][k][l] = s[i][j][k][l];
61 
62     for (int t = 2; t <= n; ++ t)
63         for (int i = 1; i <= 8; ++ i)
64             for (int j = 1; j <= 8; ++ j)
65                 for (int k = i; k <= 8; ++ k)
66                     for (int l = j; l <= 8; ++ l){
67                         if (i == k && j == l){
68                             d[t][i][j][k][l] = s[i][j][k][l];
69                             continue;
70                         }
71                         int tmp = maxint;
72                         for (int a = i; a < k; ++ a){
73                             tmp = min(tmp, d[t-1][i][j][a][l] + s[a+1][j][k][l]);
74                             tmp = min(tmp, d[t-1][a+1][j][k][l] + s[i][j][a][l]);
75                         }
76                         for (int a = j; a < l; ++ a){
77                             tmp = min(tmp, d[t-1][i][j][k][a] + s[i][a+1][k][l]);
78                             tmp = min(tmp, d[t-1][i][a+1][k][l] + s[i][j][k][a]);
79                         }
80                         d[t][i][j][k][l] = tmp;
81                     }
82 
83     double ret = (double)d[n][1][1][8][8] / n - (sum*sum/n/n);
84     ret = sqrt(ret);
85     return ret;
86 }
87 
88 int main()
89 {
90     while (scanf ("%d", &n) != EOF){
91         init();
92         printf ("%.3f\n", gao());
93     }
94     return 0;
95 }
View Code

 

posted @ 2013-11-17 20:28  Plumrain  阅读(273)  评论(4)  编辑  收藏  举报