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- 说明:这是武汉理工大学计算机学院【算法设计与分析】课程的第二次实验第一题:最大子段和问题
- >>点击查看WUTer计算机专业实验汇总
- 谨记:纸上得来终觉浅,绝知此事要躬行。
一、实验题目:
给定由n个整数组成的序列(……),求该序列形如的子段和的最大值,当所有整数均为负整数时,其最大子段和为0。
二、实验要求:
(1)分别用蛮力法、分治法和动态规划法设计最大子段和问题的算法;
(2)比较不同算法的时间性能;
(3)给出测试数据,写出程序文档。
三、实验代码
// 实验2.1最大字段和问题.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
#include <iostream>
using namespace std;
//最大字段和:蛮力法
int MaxSum_Manli(int a[], int n){
int sum, maxSum;
sum = 0;
maxSum = 0;
int i, j, k;
for (i = 0; i < n; i++) {
for (j = i; j < n; j++){
sum = 0;
for (k = i; k <= j; k++)
sum += a[k];
if (sum > maxSum)
maxSum = sum;
}
}
return maxSum;
}
//最大字段和:分治法实现
int MaxSum_Fenzhi(int a[], int left, int right) {
int sum = 0, midSum = 0, leftSum = 0, rightSum = 0, i, j;
int center, s1, s2, lefts, rights;
if (left == right) //如果序列长度为1,直接求解
sum = a[left];
else {
center = (left + right) / 2; //划分
leftSum = MaxSum_Fenzhi(a, left, center); //对应情况①,递归求解
rightSum = MaxSum_Fenzhi(a, center + 1, right); //对应情况②,递归求解
s1 = 0; lefts = 0; //以下对应情况③,先求解s1
for (i = center; i >= left; i--){
lefts += a[i];
if (lefts > s1) s1 = lefts;
}
s2 = 0; rights = 0; //再求解s2
for (j = center + 1; j <= right; j++){
rights += a[j];
if (rights > s2) s2 = rights;
}
midSum = s1 + s2; //计算情况③的最大子段和
if (midSum < leftSum) sum = leftSum; //合并解,取较大者
else sum = midSum;
if (sum < rightSum) sum = rightSum;
}
return sum;
}
//最大字段和:动态规划法
int MaxSum_Dongtai(int a[], int n){
int sum, maxSum;
int i;
sum = 0;
maxSum = 0;
for (i = 0; i < n; i++) {
sum += a[i];
if (sum > maxSum)
maxSum = sum;
else if (sum < 0)
sum = 0;
}
return maxSum;
}
int main(){
int a[100] = { 2,11,-4,13,-5,-2,20 };
int n = 7;
cout << "请输入数字串个数:";
cin >> n;
cout << "请依次输入 " << n << " 个数字:";
for (int i = 0; i < n; i++) {
cin >> a[i];
}
cout << endl;
int sum1 = MaxSum_Manli(a, n);
int sum2 = MaxSum_Fenzhi(a, 0, n-1);
int sum3 = MaxSum_Dongtai(a, n);
cout << "使用蛮力法求得:" << sum1 << endl;
cout << "使用分治法求得:" << sum2 << endl;
cout << "使用动态规划法求得:" << sum3 << endl;
}
四、运行结果: