codeforces711c Coloring Trees

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.

这个题做的我着实想吐啊，中间k不小心忘了改一下，害我检查好久

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)

注意我代码中大K和小k的区别！！！注意我代码中大K和小k的区别！！！注意我代码中大K和小k的区别！！！

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<string>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include<algorithm>using namespace std;long long dp[105][105][105];int c[205];long long p[205][205];int main(){int n,m,K;int i,j,k;scanf("%d %d %d",&n,&m,&K);memset(dp,0x3f3f3f3f3f3f3f3f,sizeof(dp));for(i=1;i<=n;i++){scanf("%d",&c[i]);}for(i=1;i<=n;i++){for(j=1;j<=m;j++){scanf("%lld",&p[i][j]);}}if(c[1]==0){for(j=1;j<=m;j++){dp[1][j][1]=p[1][j];}}else{dp[1][c[1]][1]=0;}for(i=2;i<=n;i++){if(c[i]){for(k=1;k<i;k++){for(int l=1;l<=m;l++){if(l==c[i]){dp[i][c[i]][k]=min(dp[i][c[i]][k],dp[i-1][l][k]);}else{dp[i][c[i]][k+1]=min(dp[i][c[i]][k+1],dp[i-1][l][k]);}}}}else{for(k=1;k<i;k++){for(j=1;j<=m;j++){for(int l=1;l<=m;l++){//枚举前一棵树的颜色if(l==j){//和前一棵树颜色相同dp[i][j][k]=min(dp[i-1][l][k]+p[i][j],dp[i][j][k]);}else{//和前一棵树颜色不同dp[i][j][k+1]=min(dp[i-1][l][k]+p[i][j],dp[i][j][k+1]);}}}}}}long long ans=0x3f3f3f3f3f3f3f3f;for(j=1;j<=m;j++){ans=min(ans,dp[n][j][K]);}if(ans==0x3f3f3f3f3f3f3f3f){printf("-1\n");}else{printf("%lld\n",ans);}    return 0;}