Codeforces 711c 简单dp

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 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.

吐槽：

B题看错题，认为只有0所在的行和列和对角线相等，没看Note，好伤QAQ

D题有思路了，觉得就是把环找出来，每次从环中找i条边，然后排列组合即可。没敲完QAQ

这场做的有点伤啊~~~

---

 

这题是比较简单的dp

d[i][j][k]表示第i个树涂上第j种颜色形成k组所花费的最少颜料

如果当前树已经有了颜色，那么d[i][c[i]][k]=min{ d[i-1][c[i]][k], d[i-1][j!=c[i]][k-1] }

否则d[i][j][k]=min{ d[i-1][j][k], d[i-1][x!=j][k-1] }+w[i][j]

其实想法就是如果当前涂色个上一个涂色一样，那么就d[i][j][k]由d[i-1][j][k]得到，如果不一样，由d[i-1][x!=j][k-1]得到

时间复杂度O(n^4)，优化一下可以O(n^3)

#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=(s);i<(t);i++)
#define per(i,t,s) for(int i=(t);i>=(s);i--)
#define fi first
#define se second
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int>pii;
const ll INF = 1e15 + 9;
const int N = 100 + 9;
ll d[N][N][N], w[N][N];
int c[N], n, m, K;
int main() {
    //freopen ("f.txt", "r", stdin);

    scanf ("%d%d%d", &n, &m, &K);
    rep (i, 1, n + 1) scanf ("%d", &c[i]);
    rep (i, 1, n + 1) rep (j, 1, m + 1) scanf ("%I64d", &w[i][j]);
    rep (i, 0, n + 1) rep (j, 0, m + 1) rep (k, 0, K + 1) d[i][j][k] = INF;

    d[0][0][0] = 0;
    rep (i, 1, n + 1) {
        if (c[i] != 0) {
            rep (k, 1, K + 1) {
                d[i][c[i]][k] = min (d[i][c[i]][k], d[i - 1][c[i]][k]);
                ll minn = INF;
                rep (z, 0, m + 1) if (z != c[i]) minn = min (minn, d[i - 1][z][k - 1]);
                d[i][c[i]][k] = min (d[i][c[i]][k], minn);
            }
        } else
            rep (k, 1, K + 1) {
            rep (j, 1, m + 1) {
                d[i][j][k] = min (d[i][j][k], d[i - 1][j][k] + w[i][j]);
                ll minn = INF;
                rep (z, 0, m + 1) if (z != j) minn = min (minn, d[i - 1][z][k - 1]);
                d[i][j][k] = min (d[i][j][k], minn + w[i][j]);
            }
        }
    }
    ll ans = INF;
    rep (i, 1, m + 1) ans = min (ans, d[n][i][K]);
    if (ans == INF) printf ("-1\n");
    else printf ("%I64d\n", ans);
    return 0;
}