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 }