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.
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.
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.
3 2 2
0 0 0
1 2
3 4
5 6
10
3 2 2
2 1 2
1 3
2 4
3 5
-1
3 2 2
2 0 0
1 3
2 4
3 5
5
3 2 3
2 1 2
1 3
2 4
3 5
0
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：n棵树，m种颜色，k组！求最多用m中颜色将n棵树分成k组！（只有a[i]为0才能涂色）其中每组为连续的颜色相同的数目！求分成k组的最小花费！
分析：dp[i][j][k]：表示前i棵树，分成j组，其中最后一棵树的颜色为j的最小花费！
初始化：如果第1棵树已经涂色，则第1棵数分成1组的花费为0；如果第1棵树没有涂色，则第1棵数分成1组，且颜色为m的花费为p[1][m]；其余的初始化为最大值INF，注意题目中的数据范围！
状态转移方程：如果当前树未涂色：if(a[i]==a[i-1])，则dp[i][j][k]=min(dp[i][j][k],dp[i-1][j][z]+p[i][k]);；if (a[i]!=a[i-1]),则dp[i][j+1][k]=min(dp[i][j+1][k],dp[i-1][j][k]+p[i][k]);
如果当前树已涂色：if(a[i]==a[i-1])，则dp[i][j][a[i]]=min(dp[i][j][a[i]]],dp[i-1][j][k])；if (a[i]!=a[i-1]),则dp[j][j+1][a[i]]=min(dp[i][j+1][a[i]],dp[i-1][j][k]);
AC代码：
//#include <bits/stdc++.h>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#define N 110
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios::sync_with_stdio(false)
#define INF1 0x3f3f3f3f3f3f3f3f
#define INF2 0x3f3f3f3f
#define MOD 998244353
#define Mod 1e9 + 7
template<typename T> inline T max(T a,T b,T c)
{
return max(a,max(b,c));
}
template<typename T> inline T min(T a,T b,T c)
{
return min(a,min(b,c));
}
template<typename T> inline T max(T a,T b,T c,T d)
{
return max(a,max(b,c,d));
}
template<typename T> inline T min(T a,T b,T c,T d)
{
return min(a,min(b,c,d));
}
const int dx[]= {0,1,0,-1,0,1,-1,1,-1};
const int dy[]= {0,0,1,0,-1,1,-1,-1,1};
typedef long long ll;
using namespace std;
ll p[N][N];
ll dp[N][N][N];//dp[i][x][l]:涂完前i棵树时的美丽价值为x，其中最后一棵树的颜色为l的最小代价
ll a[N];
ll n,m,k;
void init()
{
if (a[1])
dp[1][1][a[1]]=0;
else
{
for (ll i=1; i<=m; i++)
dp[1][1][i]=p[1][i];
}
}
int main()
{
scanf("%lld%lld%lld",&n,&m,&k);
memset(dp,INF2,sizeof(dp));
for (ll i=1; i<=n; i++)
scanf("%lld",&a[i]);
for (ll i=1; i<=n; i++)
for (ll j=1; j<=m; j++)
scanf("%lld",&p[i][j]);
init();
for (ll i=2; i<=n; i++)
{
if (a[i])
{
for (ll j=1; j<i; j++)
{
for (ll k=1; k<=m; k++)
{
if (k!=a[i])
dp[i][j+1][a[i]]=min(dp[i][j+1][a[i]],dp[i-1][j][k]);
else
dp[i][j][a[i]]=min(dp[i][j][a[i]],dp[i-1][j][k]);
}
}
}
else
{
for (ll j=1; j<i; j++)
{
for (ll k=1; k<=m; k++)
{
for (ll z=1; z<=m; z++)
{
if (k==z)
dp[i][j][k]=min(dp[i][j][k],dp[i-1][j][z]+p[i][k]);
else
dp[i][j+1][k]=min(dp[i][j+1][k],dp[i-1][j][z]+p[i][k]);
}
}
}
}
}
ll ans=INF1;
for (ll i=1; i<=m; i++)
ans=min(ans,dp[n][k][i]);
if (ans==INF1) printf("-1
");
else printf("%lld
",ans);
return 0;
}