【算法导论（第三版）】第四章部分习题代码

第4张 分治策略

分治问题：分解(Divide) -- 解决(Conquer) -- 合并(Conbine)

最大子数组问题 的 递归解法（O(n log n)）：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define inf (1<<29)
const int maxn = 1010;
struct node {
int left , right , sum;
node () {}
node (int left, int right, int sum) : left(left) , right(right) , sum(sum) {};
};
node find_max_crossing_subarray(int *a, int low, int mid, int high) {
int max_left , max_right;
int left_sum = - inf;
int sum = 0;
for(int i=mid;i>=low;i--) {
sum += a[i];
if(sum > left_sum) {
left_sum = sum;
max_left = i;
}
}
int right_sum = sum = 0;
for(int i=mid+1;i<=high;i++) {
sum += a[i];
if(sum > right_sum) {
right_sum = sum;
max_right = i;
}
}
return node(max_left, max_right, left_sum + right_sum);
}
node find_maximun_subarray(int *a, int low, int high) {
if(low == high) return node(low, high, a[low]);
int mid = (low + high) >> 1;
node nLeft = find_maximun_subarray(a, low, mid);
node nRight = find_maximun_subarray(a, mid+1, high);
node nCross = find_max_crossing_subarray(a, low, mid, high);
if(nLeft.sum >= nRight.sum && nLeft.sum >= nRight.sum) return nLeft;
else if(nRight.sum >= nCross.sum) return nRight;
else return nCross;
}
int n , a[maxn];
node u;
int main() {
while(scanf("%d" , &n) != EOF) {
for(int i=0;i<n;i++) scanf("%d" , &a[i]);
node ans = find_maximun_subarray(a, 0, n-1);
printf("[%d,%d] is the maximun: %d\n", ans.left, ans.right, ans.sum);
}
return 0;
}

最大子数组问题的动态规划解法（O(n)）（page.42 练习4.1-5）

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define inf (1<<29)
const int maxn = 1010;
struct node {
int left , right , sum;
node () {}
node (int left, int right, int sum) : left(left) , right(right) , sum(sum) {};
};
node find_maximun_subarray(int *a, int low, int high) {
int left, right, sum, tmp_left, tmp_right, tmp;
left = right = tmp_left = tmp_right = low;
sum = tmp = a[low];
for(int i=low+1;i<=high;i++) {
if(tmp < 0) {
tmp_left = tmp_right = i;
tmp = a[i];
} else {
tmp_right = i;
tmp += a[i];
}
if(tmp > sum) {
sum = tmp;
left = tmp_left;
right = tmp_right;
}
}
return node(left, right, sum);
}
int n , a[maxn];
node u;
int main() {
while(scanf("%d" , &n) != EOF) {
for(int i=0;i<n;i++) scanf("%d" , &a[i]);
node ans = find_maximun_subarray(a, 0, n-1);
printf("[%d,%d] is the maximun: %d\n", ans.left, ans.right, ans.sum);
}
return 0;
}

对于“矩阵乘法的Strassen算法”的一个疑问？

当矩阵的n为奇数时如何解决这个问题。