【Codeforces Round #369 (Div. 2)】Codeforces 711C Coloring Trees

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.

dp[i][j][k]表示前i棵树，第i棵的颜色是j，之前总共有k段，的最小花费。
那么考虑i-1棵树，如果颜色相同，则段数不变。否则就要新开一段。
dp[i][j][k]=min{dp[i-1][j][k],dp[i-1][x][k-1] (x!=j)}(+cost[i][j])
如果这一段是空白，j可以任取，最后还要加上费用。
否则j已经定好，不用加费用。
最后的答案是min{dp[n][x][k]}
注意边界条件和初始化。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longLL dp[110][110][110],cost[110][110];int a[110];int main(){    int i,j,k,m,n,p,q;    LL x,y,z,oo,ans;    scanf("%d%d%d",&n,&m,&k);    for (i=1;i<=n;i++)      scanf("%d",&a[i]);    for (i=1;i<=n;i++)      for (j=1;j<=m;j++)        scanf("%I64d",&cost[i][j]);    memset(dp,0x3f,sizeof(dp));    oo=dp[1][1][1];    if (a[1]) dp[1][a[1]][1]=0;    else    {        for (i=1;i<=m;i++)          dp[1][i][1]=cost[1][i];    }    for (i=2;i<=n;i++)      for (j=1;j<=m;j++)        for (p=1;p<=i&&p<=k;p++)        {            if (a[i]&&a[i]!=j) continue;            dp[i][j][p]=dp[i-1][j][p];            for (q=1;q<=m;q++)              if (q!=j)                dp[i][j][p]=min(dp[i][j][p],dp[i-1][q][p-1]);            if (!a[i]) dp[i][j][p]+=cost[i][j];        }    ans=oo;    for (i=1;i<=m;i++)      ans=min(ans,dp[n][i][k]);    if (ans==oo) printf("-1\n");    else printf("%I64d\n",ans);}