HDU 5492 Find a path (dp)

Posted 2020-08-09 Recoder

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5492

每个数字只能走下或者走右，问你方差最小多少？

方差 = (n + m - 1) * sum(a[i]^2) + sum(a[i])

我们可以定义dp[i][j][k]为(i, j)点和为k的最小平方和，k最大也就(n-m+1)*30大小

那么可以dp[i][j][k - a[i][j]] = min(dp[i][j - 1][k], dp[i - 1][j][k]) + a[i][j]*a[i][j];

保证转移最优的话，那么需要初始化dp[i][j][k] = inf;

 1 //#pragma comment(linker, "/STACK:102400000, 102400000")
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <vector>
 8 #include <cmath>
 9 #include <ctime>
10 #include <list>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long LL;
15 typedef pair <int, int> P;
16 const int N = 1e5 + 5;
17 int dp[35][35][1805], a[35][35], inf = 1e6;
18 
19 int main()
20 {
21     int t, n, m;
22     scanf("%d", &t);
23     for(int ca = 1; ca <= t; ++ca) {
24         scanf("%d %d", &n, &m);
25         for(int i = 1; i <= n; ++i) {
26             for(int j = 1; j <= m; ++j) {
27                 scanf("%d", &a[i][j]);
28                 for(int k = 0; k <= 1800; ++k) {
29                     dp[i][j][k] = inf;
30                 }
31             }
32         }
33         int sum = 0;
34         for(int i = 1; i <= n; ++i) {
35             sum += a[i][1];
36             dp[i][1][sum] = dp[i - 1][1][sum - a[i][1]] + a[i][1]*a[i][1];
37         }
38         sum = 0;
39         for(int i = 1; i <= m; ++i) {
40             sum += a[1][i];
41             dp[1][i][sum] = dp[1][i - 1][sum - a[1][i]] + a[1][i]*a[1][i];
42         }
43         for(int i = 2; i <= n; ++i) {
44             for(int j = 2; j <= m; ++j) {
45                 for(int k = a[i][j]; k <= 1800; ++k) {
46                     dp[i][j][k] = min(dp[i - 1][j][k - a[i][j]], dp[i][j - 1][k - a[i][j]]) + a[i][j]*a[i][j];
47                 }
48             }
49         }
50         int ans = inf;
51         for(int k = 0; k <= 1800; ++k) {
52             if(dp[n][m][k] < inf) {
53                 ans = min(ans, dp[n][m][k] * (n + m - 1) - k*k);
54             }
55         }
56         printf("Case #%d: %d\n", ca, ans);
57     }
58     return 0;
59 }