CF#369（Div2） C. Coloring Trees （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 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 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 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n个点，m种颜色，已经有颜色的不能涂色，没颜色为0，可以涂色，每个点i涂成颜色j花费Pij。求分成K个颜色段（112212为4个颜色段）的最小花费。无解输出-1.

解题思路：DP，dp[i][j][k] 表示枚举到第i个点有j个阶段当前颜色为k的费用。

分情况讨论，枚举到第i个点时第i个点能否上色，若不能上色，则与前一个点作比较，前

一个点与当前点颜色是否应相同，

若能够上色，则枚举所上颜色，然后再第i-1个点的各种颜色的情况作比较，取最优值。

最后答案枚举第n个点的各种颜色，取最小费用。

/* ***********************************************┆  ┏┓　　　┏┓ ┆┆┏┛┻━━━┛┻┓ ┆┆┃　　　　　　　┃ ┆┆┃　　　━　　　┃ ┆┆┃　┳┛　┗┳　┃ ┆┆┃　　　　　　　┃ ┆┆┃　　　┻　　　┃ ┆┆┗━┓　马　┏━┛ ┆┆　　┃　勒　┃　　┆　　　　　　┆　　┃　戈　┗━━━┓ ┆┆　　┃　壁　　　　　┣┓┆┆　　┃　的草泥马　　┏┛┆┆　　┗┓┓┏━┳┓┏┛ ┆┆　　　┃┫┫　┃┫┫ ┆┆　　　┗┻┛　┗┻┛ ┆************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <bitset>using namespace std;#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)#define pb push_back#define mp make_pairconst int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define LL long long#define ULL unsigned long long#define MS0(X) memset((X), 0, sizeof((X)))#define SelfType intSelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}#define Sd(X) int (X); scanf("%d", &X)#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())#define all(a) a.begin(), a.end()typedef pair<int, int> pii;typedef pair<long long, long long> pll;typedef vector<int> vi;typedef vector<long long> vll;inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")#define inf 0x3f3f3f3f3f3f3f3fvoid checkmin(LL &x,LL y){    if(x>y)x=y;}int c[105];int p[105][105];LL dp[105][105][105];  //dp[i][j][k] 表示枚举到第i个点有j个阶段当前颜色为k的费用int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);ios::sync_with_stdio(0);cin.tie(0);int n,m,K;n = read(),m = read(), K = read();for(int i=1;i<=n;i++)    {        c[i] = read();       }    for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)        {            p[i][j] = read();        }    }    memset(dp,inf,sizeof dp);    if(c[1])dp[1][1][c[1]] = 0;    else    {        for(int j=1;j<=m;j++)dp[1][1][j] = p[1][j];    }    for(int i=2;i<=n;i++)    {        if(c[i])        {            for(int j=1;j<=K;j++)            {                checkmin(dp[i][j][c[i]],dp[i-1][j][c[i]]);                for(int k=1;k<=m;k++)                {                    if(k!=c[i]) checkmin(dp[i][j][c[i]],dp[i-1][j-1][k]);                }            }        }        else        {            for(int j=1;j<=K;j++)            {                for(int k=1;k<=m;k++)                {                    checkmin(dp[i][j][k],dp[i-1][j][k]+p[i][k]);                    for(int last=1;last<=m;last++)                    {                        if(last!=k)checkmin(dp[i][j][k],dp[i-1][j-1][last]+p[i][k]);                    }                }            }        }    }    LL ans = inf;    for(int i=1;i<=m;i++)        checkmin(ans,dp[n][K][i]);    printf("%I64d\n",ans==inf?-1:ans);return 0;}