杭电 Problem-1003 Max Sum【dp】

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 200384    Accepted Submission(s): 46853

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

 

Author

Ignatius.L

思路：
    对于第i个数，它只有两种状态，一个是接在前一个队伍的前面，另一个是自己作为队头元素。dp[i]中存放的应该在i位置时i位置所属队伍的值。所以统计dp中最大的值就可以知道所有队伍中的最值。据此，只需要保证dp[i]的值最大就可以。
    so 状态转移方程为：dp[i]=max(dp[i-1]+a[i], a[i]);

 

#include <bits/stdc++.h>
#define MAXN 100005
using namespace std;
int dp[MAXN], ar[MAXN], s[MAXN];
int main() {
    int t, cnt = 0;
    scanf("%d", &t);
    while (t--) {
        int n; scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &ar[i]);
        }
        dp[1] = ar[1];
        int loc = 1;
        s[1] = 1;
        for (int i = 2; i <= n; i++) {
            dp[i] = max(dp[i-1] + ar[i], ar[i]);
            if (dp[i - 1] + ar[i] >= ar[i]) {
                s[i] = loc; dp[i] = dp[i - 1] + ar[i];
            }
            else {
                dp[i] = ar[i]; loc = s[i] = i;
            }
        }
        int ans = -1e9, e;
        for (int i = 1; i <= n; i++) {
            if (dp[i] > ans) {
                ans = dp[i]; e = i;
            }
        }
        printf("Case %d:\n%d %d %d\n", ++cnt, ans, s[e], e);
        if (t) printf("\n");
    }
    return 0;
}

 

另外的解法

#include <stdio.h>
int main()
{
	int a, b, t, m, n, Case=0, max, temp;
	scanf("%d", &t);
	while(t--) {
		Case++;
		scanf("%d", &n);
		int sum = 0;
		max = -10e5;
		temp = 1;
		for (int i = 1; i <= n; i++) {
			scanf("%d", &m);
			sum += m;
			if(sum > max) {
				b = i;
				max = sum;
				a = temp;
			}
			if(sum < 0) {
				temp = i + 1;
				sum = 0;
			}
		}
		printf("Case %d:\n", Case);
		printf("%d %d %d\n", max, a, b);
		if(t != 0) {
			printf("\n");
		}
	}
	return 0;
}