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DP
2020-04-07 |Isaac Chen | Algorithms

Ref: https://oi-wiki.org/dp/

HDU 1003

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

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.

状态转移方程:ans[i] = max( ans[i-1]+ans[i], ans[i] );

ans[i]为正时,最大区间和ans[i-1]+ans[i]
ans[i]为负,最大区间和ans[i-1]
用一个数组省内存

代码:

#include<bits/stdc++.h>
using namespace std;

int T,n,i,ans[100001],l,r,s,maxa,t=0;

int main()
{
    cin>>T;
    while(T--) {
        t++;cin>>n;
        for(i=1;i<=n;i++) scanf("%d",&ans[i]);
        maxa=ans[1];
        s=l=r=1;
        for(i=2;i<=n;i++) {
            ans[i]=max(ans[i],ans[i]+ans[i-1]);
            if(ans[i-1]<0) s=i;
            if(ans[i]>maxa) {
                l=s;
                r=i;
                maxa=ans[i];
            }
        }
        printf("Case %d:\n",t);
        printf("%d %d %d\n",maxa,l,r);    
        if(T) printf("\n");
    }
}
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