hdoj 1003 连续最大子数组 分治法 & dp法

题目大意:给出一个序列a ，寻找出a的和最大的非空连续子数组

1.使用分支策略求解 （有算导的直接看算导好了，ch4）

假设寻找的子数组A[low...high]为最大子数组，使用分支策略，意味着我们需要将原数组分为两个规模近似相同的子数组，那么我们可以找到数组a的中间位置mid

则数组A的位置可能有三种情况：

1.完全位于子数组a[low...mid] 中

2.完全位于子数组a[mid + 1... high]中

3.跨过中点

其中前两种情况我们直接使用递归即可，至于第三种情况，可以用这种策略，从中点开始，向左遍历数组求出最大的和 ，然后在从中点向右遍历，求出最大的和

之后两者相加，即获得过中点 和最大的子数组

代码(注意输出格式)

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string>#include <algorithm>#include <memory.h>using namespace std;const int maxn = 100000 + 10;const int mini = -100000000;int num[maxn];//int memo[maxn][maxn];int find_cross_subarray(int low,int high,int  & cross_low,int & cross_high)// low_max 为过中点 到达最左边的下标，high_max 为过中点 到达最右边的下标{int mid = (low + high) / 2;int res = 0;int left_sum = mini;for (int i = mid; i >= low; i--){res += num[i];if (res >= left_sum)//输出第一个 满足要求的数组 所以要尽量向左边靠拢{left_sum = res;cross_low = i;}}//for int ires = 0;int right_sum = mini;for (int i = mid + 1; i <= high; i++){res += num[i];if (res > right_sum){right_sum = res;cross_high = i;}}return right_sum + left_sum;}int find_max_subarray(int low, int high,int & max_low,int & max_high)//max_low max_high为最终答案的左右下标{if (high == low){max_low = max_high = low;return num[high];}int mid = (low + high) / 2;int left_low, left_high, left_sum;int right_low, right_high, right_sum;int cross_low, cross_high, cross_sum;left_sum = find_max_subarray(low, mid,left_low,left_high);right_sum = find_max_subarray(mid + 1, high,right_low,right_high);cross_sum = find_cross_subarray(low, high, cross_low, cross_high);if (left_sum >= cross_sum && left_sum >= right_sum){max_low = left_low;max_high = left_high;return left_sum;}if (cross_sum >= left_sum && cross_sum >= right_sum){max_low = cross_low;max_high = cross_high;return cross_sum;}max_low = right_low;max_high = right_high;return right_sum;}int main(){freopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);int casenum, length,seq = 0;cin >> casenum;while (casenum--){if (seq != 0)cout << endl;cin >> length;for (int i = 0; i < length; i++)cin >> num[i];int max_low, max_high, res;res = find_max_subarray(0, length - 1, max_low, max_high);cout << "Case " << ++seq << ":" << endl;cout << res << " " << max_low + 1<< " " << max_high + 1<<endl;}return 0;}

2.dp方法

dp[i] 表示从开始到 第i个节点连续子串的最大值 ，方程dp[i] = maxn(dp[i], dp[i - 1] + num[i])

注意需要输出第一个满足题意的答案，所以在更新最大值时注意 比较符号

代码：

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <memory.h>using namespace std;const int maxn = 100000 + 10;int num[maxn], dp[maxn];//dp[i] 表示从开始到 第i个节点连续子串的最大值 方程dp[i] = maxn(dp[i], dp[i - 1] + num[i])int max(int a, int b){return a > b ? a : b;}int main(){freopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);int casenum, length, seq = 0;cin >> casenum;while (casenum--){if (0 != seq)cout << endl;int res , end = 0, start = 0;cin >> length;for (int i = 0; i < length; i++){cin >> num[i];dp[i] = num[i];}res = dp[0];for (int i = 1; i < length; i++){dp[i] = max(dp[i], dp[i - 1] + num[i]);if (res < dp[i]){res = dp[i];end = i;}}//从end位置出发，想回找到start 位置使得 dp[i] = num[start] + ... + num[end]int temp = 0;for (int i = end; i >= 0; i--){temp += num[i];if (temp == res){start = i;//break;}}cout << "Case " << ++seq << ":"<<endl;cout << res << " " << start + 1 << " " << end + 1 << endl;}//whilereturn 0;}

另外不用dp数组方法*（实际上还有不用数组存储 数字的方法，链接：www.cnblogs.com/ACMan/archive/2012/07/01/2572061.html）：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string>using namespace std;const int maxn = 100000 + 100;int casenum;int size;int num[maxn];int ans = 0;void work(int & lp,int & rp)//给ans， lp,rp赋值{ans = num[0];int pre = ans;for (int i = 1; i < size; i++){if (pre > 0)pre += num[i];elsepre = num[i];if (pre > ans){ans = pre;rp = i;}}int tmp = 0;for (int i = rp; i >= 0; i--){tmp += num[i];if (tmp == ans)lp = i;}}int main(){freopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);int cnt = 1;cin >> casenum;while (casenum--){if (1 != cnt)cout << endl;cin >> size;for (int i = 0; i < size; i++)cin >> num[i];int lp = 0, rp = 0;work(lp, rp);cout << "Case " << cnt++ << ":" << endl;cout << ans << " " << lp + 1 << " " << rp + 1 << endl;}return 0;}