HDU1003 Max Sum【最大子段和+DP+滑动窗口法】

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 350199 Accepted Submission(s): 83372

Problem Description
Given a sequence a[1],a[2],a[3]…a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

问题链接：HDU1003 Max Sum
问题简述：（略）
问题分析：
    动态规划的的最大子段和问题，参见参考链接。这个问题跟参考链接基本上相同，只是输入输出不同，需要输出最大子段和的起始和终止序号。
    还有一种解决方法是滑动窗口法。
程序说明：（略）
参考链接：HDU1231 最大连续子序列
题记：（略）

AC的C++语言程序（滑动窗口法）如下：

/* HDU1003 Max Sum */

#include <bits/stdc++.h>

using namespace std;

const int N= 100000 + 1;
int a[N];

int main()
{
    int t, n;
    scanf("%d", &t);
    for(int k = 1; k <= t; k++) {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);

        // 滑动窗口法
        int maxSum =  INT_MIN, subSum = 0, mstart, mend, l = 1;
        for(int i = 1; i <= n; i++) {
            subSum += a[i];
            if(subSum > maxSum)
                maxSum = subSum, mstart = l, mend = i;
            if(subSum < 0)
                subSum = 0, l = i + 1;
        }

        printf("Case %d:\n", k);
        printf("%d %d %d\n", maxSum, mstart, mend);
        if(k != t) printf("\n");
    }

    return 0;
}

AC的C++语言程序如下：

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/* HDU1003 Max Sum */

#include <bits/stdc++.h>

using namespace std;

const int N= 100000;
int a[N], maxSum, mstart, mend;

void maxSubSum(int n)
{
    int subSum, sstart;
    maxSum = a[0], mstart = mend = sstart = 1;
    subSum = 0;
    for(int i = 0; i<n; i++) {
        if(subSum < 0) subSum = a[i], sstart = i + 1;
        else subSum += a[i];

        if(subSum > maxSum)
            maxSum = subSum, mstart = sstart, mend = i + 1;
    }
}

int main()
{
    int t, n;
    scanf("%d", &t);
    for(int k = 1; k <= t; k++) {
        scanf("%d", &n);
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);

        maxSubSum(n);

        printf("Case %d:\n", k);
        printf("%d %d %d\n", maxSum, mstart, mend);
        if(k != t) printf("\n");
    }

    return 0;
}

AC的C++语言程序如下：

/* HDU1003 Max Sum */

#include <bits/stdc++.h>

using namespace std;

int main()
{
    // max, mstart, mend为一组，是已经求得的最大子段和
    // sum, sstart, j为一组，是当前正在进行计算的最大子段和
    // 当前的子段不再单调增大时，则重新开启一个新的子段
    int t, n, maxSum, mstart, mend, subSum, sstart, cur;
    scanf("%d", &t);
    for(int k = 1; k <= t; k++) {
        scanf("%d", &n);

        // 一个元素时，它就是目前的最大子段和；最大子段和的起始和终止分别是maxstart和end
        scanf("%d", &cur);
        maxSum = subSum = cur;
        mstart = mend = sstart = 1;

        for(int j = 2; j <= n; j++) {
            scanf("%d", &cur);

            if(subSum < 0)
                // 不单调递增时（之前子段和为负），把当前的元素预存为另外的一个子段
                subSum = cur, sstart = j;
            else
                // 单调递增
                subSum += cur;

            // 当前正在进行计算的最大子段和超过之前的最大子段和，则重置最大子段和
            if(subSum > maxSum)
                maxSum = subSum, mstart = sstart, mend = j;
        }

        printf("Case %d:\n", k);
        printf("%d %d %d\n", maxSum, mstart, mend);
        if(k != t) printf("\n");
    }

    return 0;
}