【算法竞赛入门经典】8.1 最大连续子序列和

共提供fun1()~fun5()五种方法。

fun1()为暴力

fun2()使用了Sn进行优化，每次计算区间i~j的时候就不需要逐个累加了

fun3()是典型分治法，假设三种情况：起末位置均在mid往左，起末位置均在mid往右，起点在mid往左，终在mid往右。

fun4()是动态规划，使用了状态转移方程

sum[i] = max{sum[i-1]+dat[i],dat[i]}

fun5()是fun4()的空间优化版本，不需要开数组了。

#include<iostream>
#include<algorithm>
#include<ctime>
using namespace std;
int dat[10] = { -10,1,2,3,4,-5,-23,3,7,-21 };
int fun1() {//暴力求解
int maxi = dat[0];
for (int i = 0; i < 10; i++) {
for (int j = i; j < 10; j++) {
int sum = 0;
for (int k = i; k <= j; k++) {
sum += dat[k];
}
maxi = max(maxi, sum);
}
}
return maxi;
}
int fun2() {//sum和优化
int sum[11], maxi = dat[0];
sum[0] = 0;
for (int i = 1; i <= 10; i++)
sum[i] = dat[i - 1] + sum[i - 1];
for (int i = 0; i < 10; i++) {
for (int j = i + 1; j < 10; j++) {
int temp = sum[j] - sum[i];
maxi = max(maxi, temp);
}
}
return maxi;
}
int fun3(int left,int right) {//分治法
if (right - left == 1)
return dat[left];
int Lmax, Rmax, mid, v, sidemax;
mid = (left + right) / 2;
sidemax = max(fun3(left, mid), fun3(mid, right));
v = 0;
Lmax = dat[mid - 1];
for (int i = mid - 1; i >= left; i--) {
v += dat[i];
Lmax = max(Lmax, v);
}
v = 0;
Rmax = dat[mid];
for (int i = mid; i < right; i++) {
v += dat[i];
Rmax = max(Rmax, v);
}
return max(sidemax, Lmax + Rmax);
}
int fun4() {//dp；sum[i] = max{sum[i-1]+dat[i],dat[i]}. (sum[i]记录以a[i]为子序列末端的最大序子列连续和)
int sum[10], maxi;
sum[0] = maxi = dat[0];
for (int i = 1; i < 10; i++) {
sum[i] = max(sum[i - 1] + dat[i], dat[i]);
maxi = max(maxi, sum[i]);
}
return maxi;
}
int fun5() {//对dp的优化，不需要开数组了
int maxi, temp;
maxi = temp = dat[0];
for (int i = 1; i < 10; i++) {
if (temp > 0)
temp += dat[i];
else
temp = dat[i];
if (temp > maxi)
maxi = temp;
}
return maxi;
}
int main() {
clock_t t1, t2;
t1 = clock();
cout << fun1() << endl;
t2 = clock();
cout << t2 - t1 << " ms" << endl << endl;

t1 = clock();
cout << fun2() << endl;
t2 = clock();
cout << t2 - t1 << " ms" << endl << endl;

t1 = clock();
cout << fun3(0, 10) << endl;
t2 = clock();
cout << t2 - t1 << " ms" << endl << endl;

t1 = clock();
cout << fun4() << endl;
t2 = clock();
cout << t2 - t1 << " ms" << endl << endl;

t1 = clock();
cout << fun5() << endl;
t2 = clock();
cout << t2 - t1 << " ms" << endl << endl;

getchar();
getchar();
return 0;
}