Codeforces Round #369 (Div. 2) C. Coloring Trees DP

Posted 2020-08-11 Flowersea

C. Coloring Trees

链接：

http://codeforces.com/problemset/problem/711/C

题意：

给你nn棵树，如果cici为0的话，那么这棵树就没有上色，否则这棵树就是cici颜色的。

相同颜色的树会被当成一段，现在你要恰好刷漆刷成k段，问你最小花费是多少。

把第i棵树刷漆刷成j颜色的花费为p[i][j]

题解：

dp[i][j][k]表示第i棵树，刷成了j颜色，当前有k段的最小花费是多少。

然后好礼n^4转移就好了，很容易就能够优化成空间n^2,时间n^3的。

不优化也能过。

代码：

 1 #include<iostream>
 2 #include<algorithm>
 3 #define LL long long
 4 #define inf 1e17+1
 5 using namespace std;
 6 
 7 const int maxn = 100 + 10;
 8 LL p[maxn][maxn];
 9 LL dp[maxn][maxn][maxn];
10 int c[maxn];
11 
12 int main()
13 {
14     int n, m, kk;
15     cin >> n >> m >> kk;
16     for (int i = 1; i <= n; i++)
17         cin >> c[i];
18     for (int i = 1; i <= n; i++)
19         for (int j = 1; j <= m; j++)
20             cin >> p[i][j];
21     for (int i = 0; i <= 100; i++)
22         for (int j = 0; j <= 100; j++)
23             for (int k = 0; k <= 100; k++)
24                 dp[i][j][k] = inf;
25     dp[0][0][0] = 0;
26     for (int i = 1; i <= n; i++)
27         if (c[i] != 0)
28             for (int j = 0; j <= m; j++)
29                 for (int k = 0; k <= i; k++)
30                     if (c[i] == j)
31                         dp[i][c[i]][k] = min(dp[i - 1][j][k], dp[i][c[i]][k]);
32                     else
33                         dp[i][c[i]][k + 1] = min(dp[i - 1][j][k], dp[i][c[i]][k + 1]);
34         else
35             for (int t = 1; t <= m; t++)
36                 for (int j = 0; j <= m; j++)
37                     for (int k = 0; k <= i; k++)
38                         if (t == j)
39                             dp[i][t][k] = min(dp[i - 1][j][k] + p[i][t], dp[i][t][k]);
40                         else
41                             dp[i][t][k + 1] = min(dp[i - 1][j][k] + p[i][t], dp[i][t][k + 1]);
42     LL ans = inf;
43     for (int i = 0; i <= m; i++)
44         ans = min(dp[n][i][kk], ans);
45     if (ans == inf)
46         cout << "-1" << endl;
47     else
48         cout << ans << endl;
49     return 0;
50 }