Codeforces Round #369 (Div. 2) C. Coloring Trees （DP）

Coloring Trees

題目連接：

http://codeforces.com/contest/711/problem/C

Description

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 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.

題意：

n個樹中有部分未染色，現要使得用最小代價染色完，且數串能夠剛好劃分爲k段.

題解：

一開始想的dp，而後發現是O(n^4) (不過據說這也能過？)，後來搞了一個優化，O(n^3)過的。
dp[i][j][k]：前i個數劃分紅j段且第i個數使得第k種顏色的最小花費.
考慮轉移方程：
dp[i][j][k] = min(dp[i-1][j-1][k], min(dp[i-1][j-1][非k色])).
其中min(dp[i-1][j-1][非k色])若是單獨求解須要額外的O(n).
因此這裏額外再記錄一個dp[i][j]時使用各類顏色的最小值mfirst和次小值msecond.
在dp過程當中枚舉顏色k後，比較k與最小值mfirst的顏色是否相同，相同則取次小值，不一樣則取最小值.
這樣就將更新過程優化成O(1)了，注意已有顏色的數的處理.

代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <list>
#define LL long long
#define eps 1e-8
#define maxn 105
#define mod 100000007
#define inf 0x3f3f3f3f3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n,m, bt;
int color[maxn];
LL cost[maxn][maxn];
LL dp[maxn][maxn][maxn];
typedef pair<LL,int> pii;
pii mfirst[maxn][maxn], msecond[maxn][maxn];  /*最小值 次小值*/

int main(int argc, char const *argv[])
{
    //IN;

    while(scanf("%d %d %d" ,&n,&m,&bt) != EOF)
    {
        for(int i=1; i<=n; i++)
            scanf("%d", &color[i]);

        for(int i=1; i<=n; i++) {
            for(int j=1; j<=m; j++) {
                scanf("%I64d", &cost[i][j]);
            }
            /*已有顏色，則代價爲0，便於統一處理*/
            if(color[i]) cost[i][color[i]] = 0;
        }

        memset(dp, inf, sizeof(dp));
        memset(mfirst, inf, sizeof(mfirst));
        memset(msecond, inf, sizeof(msecond));

        for(int i=1; i<=m; i++) {
            dp[0][0][i] = 0;
        }
        mfirst[0][0] = make_pair(0, -1);
        msecond[0][0] = make_pair(0, -1);

        for(int i=1; i<=n; i++) {
            for(int j=1; j<=bt&&j<=i; j++) {
                LL mi_1 = inf, mi_2 = inf, p1, p2;
                for(int k=1; k<=m; k++) {
                    if(color[i] && k != color[i]) continue;  /*已有顏色，則只能更新該顏色*/

                    /* min(dp[i-1][j-1][非k色]) */
                    LL curmi = mfirst[i-1][j-1].first;
                    if(mfirst[i-1][j-1].second == k) curmi = msecond[i-1][j-1].first;

                    dp[i][j][k] = min(dp[i-1][j][k], curmi) + cost[i][k];
                    LL cur = dp[i][j][k];

                    if(cur < mi_1) {
                        mi_2 = mi_1; p2 = p1;
                        mi_1 = cur; p1 = k;
                    } else if(cur < mi_2) {
                        mi_2 = cur; p2 = k;
                    }
                }

                mfirst[i][j] = make_pair(mi_1, p1);
                msecond[i][j] = make_pair(mi_2, p2);
            }
        }

        LL ans = inf;
        for(int i=1; i<=m; i++) {
            ans = min(dp[n][bt][i], ans);
        }

        if(ans == inf) printf("-1\n");
        else printf("%I64d\n", ans);
    }

    return 0;
}