最大连续子串和的逐步优化

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算法渣渣。。。。没办法，从最简单的慢慢学起吧。。。哎

恩，函数基本都在20行左右~

假设全部为负数时，最大连续子串和为 0;
start, end 记录最大连续子串和的头尾。

// O(n^3)/*    三层循环， i, j 分别记录最大连续子串的头尾。*/int maxSubsequenceSum(int a[], int size, int &start, int &end){    int maxSum = 0;    for (int i = 0; i < size; ++i){        for (int j = i; j < size; ++j){            int currentSum = 0;            for (int k = i; k <= j; ++k){                currentSum += a[k];                if(currentSum > maxSum){                    maxSum = currentSum;                    start = i; end = j;                }            }        }    }    return maxSum;}

// O(n^2)/*    两层循环，sum(A[i]+...+A[j]) = A[j] + sum(A[i]+..+A[j-1]),    故可省略最里层的循环*/int maxSubsequenceSum(int a[], int size, int &start, int &end){    int maxSum = 0;    for(int i = 0; i < size; i++){        int currentSum = 0;        for(int j = i; j < size; j++){            currentSum += a[j];            if( currentSum > maxSum){                maxSum = currentSum;                start = i; end = j;            }        }    }    return maxSum;}

// O(n)
/*
一层循环，使用 startTemp 记录可能构成最大连续子串的头，

关键：若一个子序列的和是负的，则它不可能是最大连续子串的一部分。

如 { 1, -3, 4, -2, -1, 6 }， { 1, -3 }肯定不在最大连续子串中，
但 1 仍可能是最大子序列，注意保存该情况。
*/

int maxSubsequenceSum(int a[], int size, int &start, int &end){    int maxSum = 0;    int startTemp = 0;    int currentSum = 0;    start = end = 0;    for(int i = 0; i < size; i++){        currentSum += a[i];        if(currentSum < 0){            startTemp = i + 1;            currentSum = 0;        }        if(currentSum > maxSum){            maxSum = currentSum;            start = startTemp;            end = i;        }    }    return maxSum;}

这里随便找了一道例题：HDU 1003，注：此处需考虑全为负数情况应选最大的那个负数。

#include<iostream>#include<cstdio>using namespace std;const int maxn=100005;int a[maxn];int maxSubsequenceSum(int a[], int &size, int &start, int &end){    int maxSum = 0;    int startTemp = 0;    int currentSum = 0;    for(int i = 0; i < size; i++){        currentSum += a[i];        if(currentSum < 0){            startTemp = i + 1;            currentSum = 0;        }        if(currentSum > maxSum){            maxSum = currentSum;            start = startTemp;            end = i;        }    }    return maxSum;}int main(){int t, n;cin >> t;for(int m = 1; m <= t; m++){cin >> n;int tag = 0;for(int i = 0; i < n; i++){scanf("%d", &a[i]);if(a[i] > 0)tag = 1;}cout << "Case " << m << ":" << endl;int start = 0, end = 0, ans;if(tag)//含有正数的情况ans = maxSubsequenceSum(a, n, start, end);else{//考虑全部非正的特殊情况ans = a[0];for(int i = 1; i < n; i++){if(a[i] > ans){ans = a[i];start = i; end = i;}}}cout << ans <<" "<< start + 1 <<" "<< end + 1 << endl;if(m != t)cout << endl;}return 0;}