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Codeforces Round #369 (Div. 2) C. Coloring Trees 数位dp,好题

2016-09-08 09:34:44         来源:ProLights的博客  
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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 wherentrees grow. They decided to be naughty and color the trees in the park. The trees are numbered with integers from1tonfrom left to right.

Initially, treeihas colorci. ZS the Coder and Chris the Baboon recognizes onlymdifferent colors, so0?≤?ci?≤?m, whereci?=?0means that treeiisuncolored.

ZS the Coder and Chris the Baboon decides to color only the uncolored trees, i.e. the trees withci?=?0. They can color each of them them in any of themcolors from1tom. Coloring thei-th tree with colorjrequires exactlypi,?jlitres of paint.

The two friends define thebeautyof a coloring of the trees as theminimumnumber of contiguous groups (each group contains some subsegment of trees) you can split all thentrees into so that each group contains trees of the same color. For example, if the colors of the trees from left to right are2,?1,?1,?1,?3,?2,?2,?3,?1,?3, the beauty of the coloring is7, since we can partition the trees into7contiguous 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 isexactlyk. 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,mandk(1?≤?k?≤?n?≤?100,1?≤?m?≤?100)— the number of trees, number of colors and beauty of the resulting coloring respectively.

The second line containsnintegersc1,?c2,?...,?cn(0?≤?ci?≤?m), the initial colors of the trees.ciequals to0if the tree numberiis uncolored, otherwise thei-th tree has colorci.

Thennlines follow. Each of them containsmintegers. Thej-th number on thei-th of them line denotespi,?j(1?≤?pi,?j?≤?109)— the amount of litres the friends need to colori-th tree with colorj.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 beautyk, 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 colors2,?1,?1minimizes the amount of paint used, which equals to2?+?3?+?5?=?10. Note that1,?1,?1would not be valid because the beauty of such coloring equals to1({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 is3, 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 matchesk, so no paint is used and the answer is0.


 

 

Source

Codeforces Round #369 (Div. 2)

 

My Solution

数位dp, 好题

状态定义 dp i j k 为 当前在 从左往右 第 i 位, 且 以 j 结尾, 已经有 k 个片段

初始化 dp[ i ][ j ][ k ] = INF

边界 if(i == 1) ......

状态转移方程 if(val[ i ] == 0) dp[ i ][ j ][ k ] = min(dp[ i ][ j ][ k ], dp[ i - 1 ][ x ][ k - 1] + p[ i ][ j ]); // x = 1 ~ m, x != j

else dp[ i ][ j ][ k ] = min(dp[ i ][ j ][ k ], dp[ i - 1][ x ][ k - 1]); // x = 1 ~ m, x != j

复杂度 O(n^4)

 

 

#include 
#include 
#include 

using namespace std;
typedef long long LL;
const int maxn = 1e2 + 8;


LL col[maxn], p[maxn][maxn], dp[maxn][maxn][maxn];

int main()
{
    #ifdef LOCAL
    freopen("c.txt", "r", stdin);
    //freopen("o.txt", "w", stdout);
    int T = 4;
    while(T--){
    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);


    LL n, m, kk;
    cin >> n >> m >> kk;
    for(int i = 1; i <= n; i++){
        cin >> col[i];
    }
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            cin >> p[i][j];
        }
    }

    for(int i = 0; i <= n + 1; i++){
        for(int  j = 0; j <= m + 1; j++){
            for(int k = 0; k <= kk + 1; k++){
                dp[i][j][k] = 9e18;
            }
        }
    }
    //cout< i) break;
            if(col[i] != 0){
                if(i == 1){
                    dp[i][col[i]][1] = 0;
                    continue;
                }
                j = col[i];
                dp[i][j][k] = dp[i-1][j][k];
                for(y = 1; y <= m; y++){
                    if(j == y) continue;
                    else dp[i][j][k] = min(dp[i][j][k], dp[i - 1][y][k - 1]);
                }
            }
            else{
                if(i == 1){
                    for(y = 1; y <= m; y++){
                        dp[i][y][1] = p[1][y];
                    }
                    continue;
                }

                for(j = 1; j <= m; j++){
                    dp[i][j][k] = dp[i - 1][j][k] + p[i][j];
                    for(y = 1; y <= m; y++){
                        if(j == y) continue;
                        else dp[i][j][k] = min(dp[i][j][k], dp[i - 1][y][k - 1] + p[i][j]);
                    }
                }
            }
        }
    }


    LL ans = 9e18;
    for(j = 1; j <= m; j++){
        ans = min(ans, dp[n][j][kk]);
    }
    if(ans == 9e18) cout << -1 << endl;
    else cout << ans << endl;


    #ifdef LOCAL
    cout << endl;
    }
    #endif // LOCAL
    return 0;
}
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