数组分段和最大值最小问题

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1. 题目：给定一个数组，和一个值k，数组分成k段。要求这k段子段和最大值最小。求出这个值。

2.分析：这道题目很经典，也很难，个人认为很难。文章中给出了三种算法：算法1，暴力搜索。本题暴力搜索算法并不是很明显，可以使用递归实现暴力搜索。递归首先要有递归式：

n n-1

M[n, k] = min { max { M[j, k-1], ∑ Ai } }

j=1 i=j

n表示数组长度，k表示数组分成几段。初始化条件：

M[1, k] = A0

n-1
M[n, 1] = ∑ Ai

i=0

View Code
#include <iostream>
#include <cassert>

using namespace std;
int sum(int A[], int from, int to) {
int total = 0;
for (int i = from; i <= to; i++)
total += A[i];
return total;
}
//递归的暴力搜素算法
//指数时间的复杂度
int partition(int A[], int n, int k) {
if (k == 1)
return sum(A, 0, n-1);
if (n == 1)
return A[0];

int best = INT_MAX;
for (int j = 1; j <= n; j++)
best = min(best, max(partition(A, j, k-1), sum(A, j, n-1)));

return best;
}

//改进的动态规划算法
//时间复杂度：O(kN2)
//空间复杂度：O(kN)
const int MAX_N = 100;
int findMax(int A[], int n, int k) {
int M[MAX_N+1][MAX_N+1] = {0};
int cum[MAX_N+1] = {0};
for (int i = 1; i <= n; i++)
cum[i] = cum[i-1] + A[i-1];

for (int i = 1; i <= n; i++)
M[i][1] = cum[i];
for (int i = 1; i <= k; i++)
M[1][i] = A[0];

for (int i = 2; i <= k; i++) {
for (int j = 2; j <= n; j++) {
int best = INT_MAX;
for (int p = 1; p <= j; p++) {
best = min(best, max(M[p][i-1], cum[j]-cum[p]));
}
M[j][i] = best;
}
}
return M
[k];
}

int getMax(int A[], int n) {
int max = INT_MIN;
for (int i = 0; i < n; i++) {
if (A[i] > max) max = A[i];
}
return max;
}

int getSum(int A[], int n) {
int total = 0;
for (int i = 0; i < n; i++)
total += A[i];
return total;
}

int getRequiredPainters(int A[], int n, int maxLengthPerPainter) {
int total = 0, numPainters = 1;
for (int i = 0; i < n; i++) {
total += A[i];
if (total > maxLengthPerPainter) {
total = A[i];
numPainters++;
}
}
return numPainters;
}

//想不到的二分查找算法
//时间复杂度：O(N log ( ∑ Ai )).
//空间复杂度：0(1)
int BinarySearch(int A[], int n, int k) {
int lo = getMax(A, n);
int hi = getSum(A, n);

while (lo < hi) {
int mid = lo + (hi-lo)/2;
int requiredPainters = getRequiredPainters(A, n, mid);
if (requiredPainters <= k)
hi = mid;
else
lo = mid+1;
}
return lo;
}
int main()
{
enum{length=9};
int k=3;
int a[length]={9,4,5,12,3,5,8,11,0};
cout<<partition(a,length,k)<<endl;
cout<<findMax(a,length,k)<<endl;
cout<<BinarySearch(a,length,k)<<endl;
return 0;
}

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