codeforces 369 div2 C dp

http://codeforces.com/contest/711

C. 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 ci. ZS the Coder and Chris the Baboon recognizes only m different colors, so 0 ≤ ci ≤ m, where ci = 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 ci = 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 pi, 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, ..., cn (0 ≤ ci ≤ m), the initial colors of the trees. ci equals to 0 if the tree number i is uncolored, otherwise the i-th tree has color ci.

Then n lines follow. Each of them contains m integers. The j-th number on the i-th of them line denotes pi, j (1 ≤ pi, j ≤ 109) — the amount of litres the friends need to color i-th tree with color j. pi, 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.

题目大意：给你一个串，串上的每个点都有颜色（颜色为0~m），0表示无色。目前，我们要给无色的点涂色，给第i个串图上第j中颜色需要花费P[i][j]升油料。我们定义串的colorful为颜色不连续的点的个数。现在，我们要求串的colorful为k，问所需要的最小的颜料花费是多少？

思路：定义dp[i][j][k]表示第i棵点涂上第j种颜色在目前colorful在k的条件下的最小使用量

//看看会不会爆int! 或者绝对值问题。
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define all(a) a.begin(), a.end()
const int maxn =  + ;
const LL inf = 1e18;
LL p[maxn][maxn], dp[maxn][maxn][maxn];
int n, m, ty;
int c[maxn];

int main(){
    cin >> n >> m >> ty;
    for (int i = ; i <= n; i++) scanf("%d", c + i);
    for (int i = ; i <= n; i++){
        for (int j = ; j <= m; j++){
            scanf("%I64d", &p[i][j]);
        }
    }
    for (int i = ; i <= n; i++){
        for (int j = ; j <= m; j++){
            for (int k = ; k <= n; k++){
                dp[i][j][k] = inf;
            }
        }
    }
    for (int i = ; i <= n; i++){
        if (c[i] != ){
            for (int j = ; j <= m; j++){//前一个的color
                for (int k = ; k <= i; k++){
                    if (k == ){
                        dp[i][c[i]][k] = min(dp[i][c[i]][k], dp[i - ][c[i]][k]);
                    }
                    else {
                        if (c[i] == j) dp[i][c[i]][k] = min(dp[i - ][j][k], dp[i][c[i]][k]);
                        else dp[i][c[i]][k] = min(dp[i - ][j][k - ], dp[i][c[i]][k]);
                    }
                    //printf("%d%c", dp[i])
                }
            }
        }
        else {
            for (int j = ; j <= m; j++){//目前的color
                for (int k = ; k <= i; k++){
                    if (k == ){//至少有一种颜色
                        dp[i][j][k] = min(dp[i][j][k], dp[i - ][j][k] + p[i][j]);
                    }
                    else {
                         dp[i][j][k] = min(dp[i - ][j][k] + p[i][j], dp[i][j][k]);//加入颜色相等
                         LL minval = inf;
                         for (int c = ; c <= m; c++){
                             if (c == j) continue;
                             minval = min(minval, dp[i - ][c][k - ] + p[i][j]);
                         }
                         dp[i][j][k] = min(dp[i][j][k], minval);
                    }
                }
            }
        }
    }
    LL ans = inf;
    for (int i = ; i <= m; i++){
        ans = min(ans, dp[n][i][ty]);
    }
    if (ans == inf) ans = -;
    printf("%I64d\n", ans);
    return ;
}