CodeForce 711C -- Coloring Trees

Coloring Trees

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

　　ZS the Coder and Chris the Baboon has arrived at Udayland! They walked in the park where n trees grow. They decided to be naughty and color the trees in the park. The trees are numbered with integers from 1 to n from left to right.

Initially, tree i has color c_i. ZS the Coder and Chris the Baboon recognizes only m different colors, so 0 ≤ c_i ≤ m, where c_i = 0 means that tree i is uncolored.

ZS the Coder and Chris the Baboon decides to color only the uncolored trees, i.e. the trees with c_i = 0. They can color each of them them in any of the m colors from 1 to m. Coloring the i-th tree with color j requires exactly p_i, j litres of paint.

The two friends define the beauty of a coloring of the trees as the minimum number of contiguous groups (each group contains some subsegment of trees) you can split all the n trees into so that each group contains trees of the same color. For example, if the colors of the trees from left to right are 2, 1, 1, 1, 3, 2, 2, 3, 1, 3, the beauty of the coloring is 7, since we can partition the trees into 7 contiguous groups of the same color : {2}, {1, 1, 1}, {3}, {2, 2}, {3}, {1}, {3}.

ZS the Coder and Chris the Baboon wants to color all uncolored trees so that the beauty of the coloring is exactly k. They need your help to determine the minimum amount of paint (in litres) needed to finish the job.

Please note that the friends can't color the trees that are already colored.

Input

 

　　The first line contains three integers, n, m and k (1 ≤ k ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of trees, number of colors and beauty of the resulting coloring respectively.

The second line contains n integers c₁, c₂, ..., c_n (0 ≤ c_i ≤ m), the initial colors of the trees. c_i equals to 0 if the tree number i is uncolored, otherwise the i-th tree has color c_i.

Then n lines follow. Each of them contains m integers. The j-th number on the i-th of them line denotes p_i, j (1 ≤ p_i, j ≤ 10⁹) — the amount of litres the friends need to color i-th tree with color j. p_i, j's are specified even for the initially colored trees, but such trees still can't be colored.

 

Output

　　Print a single integer, the minimum amount of paint needed to color the trees. If there are no valid tree colorings of beauty k, print  - 1.

Examples

 

Input

 

3 2 2
0 0 0
1 2
3 4
5 6

Output

 

10

Input

 

3 2 2
2 1 2
1 3
2 4
3 5

Output

 

-1

Input

 

3 2 2
2 0 0
1 3
2 4
3 5

Output

 

5

Input

 

3 2 3
2 1 2
1 3
2 4
3 5

Output

 

0

Note

　　In the first sample case, coloring the trees with colors 2, 1, 1 minimizes the amount of paint used, which equals to 2 + 3 + 5 = 10. Note that 1, 1, 1 would not be valid because the beauty of such coloring equals to 1 ({1, 1, 1} is a way to group the trees into a single group of the same color).

In the second sample case, all the trees are colored, but the beauty of the coloring is 3, so there is no valid coloring, and the answer is  - 1.

In the last sample case, all the trees are colored and the beauty of the coloring matches k, so no paint is used and the answer is 0.

 

题意：

　　给你n颗树，每棵树可以涂m种颜色，把树分成k组，连续相同颜色的树为合并为一组。

　　给出了n颗树的涂色情况，当涂色为0的时候，表示还没有涂颜色，

　　给n * m 的矩阵，第i行j列为 第i个点，图j颜色的消耗。

　　求n颗树涂颜色涂成k组时，用的最少消耗是多少，

分析：

　　dp[i][k][j] 表示到了第i个点，分成k组， 颜色为j时，最少消耗为多少；

　　当前点被涂过颜色：

　　则 dp[now][j + (a[i] != k)][a[i]] = min(dp[now][j + (a[i] != k)][a[i]], dp[next][j][k]);

　　当没被涂过：

　　枚举颜色：

　　　　for(int l = 1; l <= m; l++)
        　　dp[now][j + (l != k)][l] = min(dp[now][j + (l != k)][l], dp[next][j][k] + val[i][l]);

AC代码：

# include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
const int MAX = 110; 
const ll INF = (1LL << 60);
ll a[MAX], val[MAX][MAX], dp[2][MAX][MAX];
int main()
{
    int n, m, zu;
    scanf("%d%d%d", &n, &m, &zu);
    for(int i = 1; i <= n; i++)
        scanf("%I64d", &a[i]);

    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            scanf("%I64d", &val[i][j]);

    int now = 0, next = 1;
    memset(dp[now], 0x3f3f3f3f, sizeof(dp[now]));
    dp[now][0][0] = 0;
    for(int i = 1; i <= n; i++)
    {
        swap(now, next);
        memset(dp[now], 0x3f3f3f3f, sizeof(dp[now]));
        
        for(int j = 0; j <= n; j++)
            for(int k = 0; k <= m; k++)
            {
                if(a[i])
                    dp[now][j + (a[i] != k)][a[i]] = min(dp[now][j + (a[i] != k)][a[i]], dp[next][j][k]);
                else
                    for(int l = 1; l <= m; l++)
                        dp[now][j + (l != k)][l] = min(dp[now][j + (l != k)][l], dp[next][j][k] + val[i][l]);
            }
    }
    ll Min = INF;
    for(int i = 0; i <= m; i++)
        Min = min(dp[now][zu][i], Min);
    
    printf("%I64d\n", Min < INF ? Min : -1);
    return 0;
}