Consider a grid of size 1 x N. Each cell of the grid has the following properties

·         Cell C of the grid has a value of V_C (1 ≤ C ≤ N)

·         The value of each cell is a positive integer less than 26

·         Some of the cells are special and they are represented with the character X

·         Cell C has a weight of (two to the power of cell value)

·         The special cells have weights of 1

You will be given the values of these N cells and your job will be to divide these into K segments so that

·         Each segment contains at least one cell

·         There is at least one special cell in each segment

The weight of a segment is equal to the product of the weights of the cells it contains. You have to form segments in such a way so that ratio

(Highest weight of all the segments) / (Lowest weight of all the segments) is minimized.

In case there are multiple answers with the same lowest ratio, you have to make sure the number of cells in the first segment is maximized. If there is still a tie, then make sure the number of cells in the second segment is maximized and so on.

Example:

N = 5 and K = 2
Cell values = {1 2 X 3 X }
Cell weights = {2 4 1 8 1}
Optimal segmentation = (2 4 1)(8 1)
Weights of segments = (8)(8)
Ratio = 1
Final Result = (1 2 X)(3 X)

Input

The first line of input is an integer T(T ≤ 200) that indicates the number of test cases. Each case starts with two integers N(1 < N < 31) and K(1 < K < 16). The meaning of N and K are mentioned above. The next line contains N integers where the I^th integer gives the value of V_I. The integers that are special will be represented by X.

Output

For each case, output the case number first. If there is no way to divide the N cells into K segments, meeting the constraints above, then print “not possible!” If there is a way but the ratio is greater than 10¹⁵ then print “overflow”. If the ratio is not greater than 10¹⁵ then output the ratio first followed by the segmentations. Each segment is enclosed by brackets. Look at the output for detailed format.

Sample Input                            Output for Sample Input

4 X 3 1 X 3 X X 3 2

Problemsetter: Sohel Hafiz

Special Thanks: Md. Arifuzzaman Arif

題目描述：

將 N 個數字分成 K 段，每一段都要包含一個 X。

求分段後，抓每一段的最大值 - 最小值最小化 (最小化最大差額)。

最後要輸出解的分法，按照字典順序，越左邊的段個數越多越好。

題目解法：

狀態 dp[i][j][k] 討論前 i 個，切成 j 段，其中最小值為 k 的最小最大值為何。

最麻煩就是輸出解法了，字典順序搞了好久才找到輸出的方法。

#include <stdio.h>
#include <string.h>
#include <set>
#include <vector>
#include <algorithm>
using namespace std;
int V[35], N, K;
int dp[31][16][30*26 + 5]; // [i-th][segment_count][min_value]
int ac_dp[31][16][30*26 + 5];
void solve() {
    int i, j, k, prev;
    int sum[35] =  {}, sumX[35] = {}, nextX[35] = {};
    for(i = 1; i <= N; i++) {
        sum[i] = sum[i-1] + V[i];
        sumX[i] = sumX[i-1] + (V[i] == 0);
    }
    for(i = N, prev = N+1; i >= 0; i--) {
        nextX[i] = prev;
        if(V[i] == 0)
            prev = i;
    }
    if(sumX[N] < K) {
        puts("not possible!");
        return ;
    }
    set<int> R[31][16];
    dp[0][0][0] = 0, R[0][0].insert(0);
    for(i = 0; i < N; i++) {
        for(j = 0; j < K; j++) {
            for(set<int>::iterator it = R[i][j].begin();
                it != R[i][j].end(); it++) {
                for(k = nextX[i]; k <= N; k++) {
                    int s = sum[k] - sum[i];
                    int mn = min(*it, s), mx = max(dp[i][j][*it], s);
                    if(i == 0)    mn = mx = s;
                    if(R[k][j+1].find(mn) == R[k][j+1].end())
                        R[k][j+1].insert(mn), dp[k][j+1][mn] = 0xfffffff, ac_dp[k][j+1][mn] = 0;
                    dp[k][j+1][mn] = min(dp[k][j+1][mn], mx);
                }
            }
        }
    }
    int ret = 0xfffffff;
    for(set<int>::iterator it = R[N][K].begin();
        it != R[N][K].end(); it++)  {
        ret = min(ret, dp[N][K][*it] - *it);
    }
    if(ret >= 50) {
        puts("overflow");
        return;
    }
    printf("%lld ", 1LL<<ret);
    vector< pair<int, int> > P;
    for(set<int>::iterator it = R[N][K].begin();
        it != R[N][K].end(); it++)  {
        if(ret == dp[N][K][*it] - *it) {
            ac_dp[N][K][*it] = 1;
            P.push_back(make_pair(*it, dp[N][K][*it]));
        }
    }   
    for(i = N; i >= 0; i--) {
        for(j = 0; j < K; j++) {
            for(set<int>::iterator it = R[i][j].begin();
                it != R[i][j].end(); it++) {
                for(k = nextX[i]; k <= N; k++) {
                    int s = sum[k] - sum[i];
                    int mn = min(*it, s), mx = max(dp[i][j][*it], s);
                    if(i == 0)    mn = mx = s;
                    if(ac_dp[k][j+1][mn]) {
                        int ok = 0;
                        for(int p = 0; p < P.size(); p++) {
                            if(*it >= P[p].first && dp[i][j][*it] <= P[p].second)
                                ok = 1;
                        }
                        if(ok) {
                            ac_dp[i][j][*it] = ok;
                            break;
                        }
                    }
                }
            }
        }
    }
    int idx = 0, idx_mn = 0;
    for(int seg = 0; seg < K; seg++) {
        for(j = N; j >= idx; j--) {
            if(sumX[j] - sumX[idx] == 0)
                continue;
            int s = sum[j] - sum[idx];
            int mn = min(idx_mn, s), mx = max(dp[idx][seg][idx_mn], s);
            if(idx == 0)    mn = mx = s;
            if(ac_dp[j][seg+1][mn]) {
                putchar('(');
                for(k = idx+1; k <= j; k++) {
                    if(V[k])
                        printf("%d", V[k]);
                    else
                        printf("X");
                    printf("%c", k == j ? ')' : ' ');
                }
                idx = j, idx_mn = mn;
                break;   
            }
        }
    }
    puts("");
}
int main() {
    int testcase, cases = 0;
    int i, j, k;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d", &N, &K);
        for(i = 1; i <= N; i++) {
            char s[10];
            scanf("%s", s);
            if(s[0] == 'X')
                V[i] = 0;
            else
                sscanf(s, "%d", &V[i]);
        }
        printf("Case %d: ", ++cases);
        solve();
    }
    return 0;
}
/*
1000
20 6
X 3 7 X 2 4 6 3 X 6 9 4 X X 4 7 X X 5 6
20 12
X X 2 X 7 2 X X X X 1 X X 9 X X X X 4 X
*/