C | Optimal Segments Input: Standard Input Output: Standard Output | 
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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 VC (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 Ith integer gives the value of VI. 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 5 2 1 2 X 3 X 6 3 X X 2 3 4 5 10 3 X X X 25 25 25 25 25 25 25 10 3 4 X 3 1 X 3 X X 3 2 | Case 1: 1 (1 2 X)(3 X) Case 2: not possible! Case 3: overflow Case 4: 8 (4 X 3)(1 X 3 X)(X 3 2) |
Problemsetter: Sohel Hafiz
Special Thanks: Md. Arifuzzaman Arif
目描述：
N 字分成 K 段，每一段都要包含一 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)。
最後要出解的分法，按照字典序，越左的段越多越好。
目解法：
dp[i][j][k] 前 i ，切成 j 段，其中最小值 k 的最小最大值何。
最麻就是出解法了，字典序搞了好久才找到出的方法。
#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;
&nbp; 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
*/
