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

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：

#include<iostream>
#include<algorithm>
#define LL long long
#define inf 1e17+1
using namespace std;

const int maxn = 100 + 10;
LL p[maxn][maxn];
LL dp[maxn][maxn][maxn];
int c[maxn];

int main()
{
    int n, m, kk;
    cin >> n >> m >> kk;
    for (int i = 1; i <= n; i++)
        cin >> c[i];
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> p[i][j];
    for (int i = 0; i <= 100; i++)
        for (int j = 0; j <= 100; j++)
            for (int k = 0; k <= 100; k++)
                dp[i][j][k] = inf;
    dp[0][0][0] = 0;
    for (int i = 1; i <= n; i++)
        if (c[i] != 0)
            for (int j = 0; j <= m; j++)
                for (int k = 0; k <= i; k++)
                    if (c[i] == j)
                        dp[i][c[i]][k] = min(dp[i - 1][j][k], dp[i][c[i]][k]);
                    else
                        dp[i][c[i]][k + 1] = min(dp[i - 1][j][k], dp[i][c[i]][k + 1]);
        else
            for (int t = 1; t <= m; t++)
                for (int j = 0; j <= m; j++)
                    for (int k = 0; k <= i; k++)
                        if (t == j)
                            dp[i][t][k] = min(dp[i - 1][j][k] + p[i][t], dp[i][t][k]);
                        else
                            dp[i][t][k + 1] = min(dp[i - 1][j][k] + p[i][t], dp[i][t][k + 1]);
    LL ans = inf;
    for (int i = 0; i <= m; i++)
        ans = min(dp[n][i][kk], ans);
    if (ans == inf)
        cout << "-1" << endl;
    else
        cout << ans << endl;
    return 0;
}