Codeforces Round #369 (Div. 2) [C] 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 wheren trees grow. They decided to be naughty and color the trees in the park. The trees are numbered with integers from1 to n from left to right.

Initially, tree i has colorc_i. ZS the Coder and Chris the Baboon recognizes onlym different colors, so 0 ≤ c_i ≤ m, wherec_i = 0 means that treei is uncolored.

ZS the Coder and Chris the Baboon decides to color only the uncolored trees, i.e. the trees withc_i = 0. They can color each of them them in any of them colors from 1 tom. Coloring the i-th tree with colorj 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 then trees into so that each group contains trees of the same color. For example, if the colors of the trees from left to right are2, 1, 1, 1, 3, 2, 2, 3, 1, 3, the beauty of the coloring is7, 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 isexactly 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 integersc₁, c₂, ..., c_n (0 ≤ c_i ≤ m), the initial colors of the trees. c_i equals to0 if the tree number i is uncolored, otherwise thei-th tree has color c_i.

Then n lines follow. Each of them containsm integers. The j-th number on thei-th of them line denotes p_i, j (1 ≤ p_i, j ≤ 10⁹) — the amount of litres the friends need to colori-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 beautyk, print  - 1.

Examples

Input

3 2 20 0 01 23 45 6

Output

10

Input

3 2 22 1 21 32 43 5

Output

-1

Input

3 2 22 0 01 32 43 5

Output

5

Input

3 2 32 1 21 32 43 5

Output

0

Note

In the first sample case, coloring the trees with colors2, 1, 1 minimizes the amount of paint used, which equals to2 + 3 + 5 = 10. Note that 1, 1, 1 would not be valid because the beauty of such coloring equals to1 ({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 is3, 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 matchesk, so no paint is used and the answer is 0.

题意:给你一个序列，每一个数字都在[0,m]范围内，每一个为0的数字都可以换成另外一个[1,m]的数字

将第i个数字换成j的代价为cost[i][j]，求出让整个序列的段数为K的最小代价(所有的0都必须清除)

f[i][j][k]表示讨论第i个数字之后已经讨论的序列的最后一个数为j并且整个序列已经分为k段的最小代价

方程在下面代码里

#include<cstdio>#include<iostream>#define LL long longusing namespace std;const LL inf=1e18;LL n,m,K,s[105],cost[105][105];LL f[105][105][105];int main(){scanf("%d%d%d",&n,&m,&K);LL i,j,k,c,ans=inf;for(i=1;i<=n;i++)scanf("%d",&s[i]);for(i=1;i<=n;i++)    for(j=1;j<=m;j++)        scanf("%d",&cost[i][j]);for(i=1;i<=n;i++)    for(j=1;j<=m;j++)        for(k=0;k<=n;k++)            f[i][j][k]=inf;for(i=1;i<=n;i++)    for(k=1;k<=i;k++)        if(s[i]){        f[i][s[i]][k]=f[i-1][s[i]][k];        for(j=1;j<=m;j++)            if(s[i]!=j)                f[i][s[i]][k]=min(f[i][s[i]][k],f[i-1][j][k-1]);}else {for(j=1;j<=m;j++){for(c=1;c<=m;c++){if(j==c)f[i][c][k]=min(f[i][c][k],f[i-1][j][k]+cost[i][c]);else f[i][c][k]=min(f[i][c][k],f[i-1][j][k-1]+cost[i][c]);}}}for(i=1;i<=m;i++)    ans=min(f[n][i][K],ans);if(ans>=inf)cout<<-1;else cout<<ans; }