HDU1003Max Sum(original) HDU1024Max Sum Plus Plus(upgrading)

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 180760    Accepted Submission(s): 42247

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

最简单版本的最大连续子序列和，无需多说，直接贪心。注意全是负数的情况即可。

#include<cstdio>#include<algorithm>#define INF 1<<30using namespace std;int a[100005];int main(){    int t,tc = 0, _t;    scanf("%d",&t);    _t = t - 1;    while(t--){        int n;        scanf("%d",&n);        if(t != _t)printf("\n");        int tag = 0;        int sum = 0, ans = 0, tmp = 0,l = 0,r = 0;        int maxnum = -INF, index;        for(int i = 0; i < n; i++){            scanf("%d",a+i);            if(a[i]>0)tag = 1;            if(maxnum < a[i])maxnum = a[i], index = i;            sum += a[i];            if(sum < 0) sum = 0,tmp = i+1;            if(ans < sum){                l = tmp;                r = i;                ans = sum;            }        }        if(!tag){            printf("Case %d:\n%d %d %d\n",++tc, maxnum, index+1,index+1);            continue;        }        printf("Case %d:\n%d %d %d\n",++tc,ans,l+1,r+1);    }    return 0;}

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20804    Accepted Submission(s): 6925

Problem Description

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S₁, S₂, S₃, S₄ ... S_x, ... S_n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S_x ≤ 32767). We define a function sum(i, j) = S_i + ... + S_j (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + sum(i_m, j_m) maximal (i_x ≤ i_y ≤ j_x or i_x ≤ j_y ≤ j_x is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i_x, j_x)(1 ≤ x ≤ m) instead. ^_^

 

Input

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

 

Output

Output the maximal summation described above in one line.

 

Sample Input

1 3 1 2 32 6 -1 4 -2 3 -2 3

升级版，题意大概可以表述成：要求将给定串分成若干不相交的部分，求各部分的max sum的最大和。

思路：动态规划。dp[ i ][ j ]表示前 j 个元素分成 i 段得到的最大和，则有状态转移方程dp[ i ][ j ]=max(dp[ i ][ j-1], dp[ i-1 ][ k ]) + a[ j ]，k: i-1 ->  j-1，表示第 j 个元素的两种情况：加入前一个连续子串和自己重新成为一个串。那么题目大概就差不多了，注意优化成一维的形式（或者滚动数组），并且为了减小复杂度，要开一个数组记录i - 1行的dp值。

#include<cstdio>const int maxn = 1000005, INF = 1<<30;int pre[maxn], dp[maxn], a[maxn], maxnum,m ,n;int main(){    while(~scanf("%d %d",&m,&n)){        for(int i = 1; i <= n; i++) scanf("%d",a+i),dp[i]=pre[i]=0;        dp[0]=pre[0]=0;        for(int i = 1; i <= m; i++){            maxnum = -INF;            for(int j = i; j <= n; j++){                dp[j] = (dp[j-1]>pre[j-1]?dp[j-1]:pre[j-1])+a[j];                pre[j-1] = maxnum;                maxnum = maxnum>dp[j]?maxnum:dp[j];            }        }        printf("%d\n",maxnum);    }    return 0;}