POJ2479 Maximum sum[DP|最大子段和]
Posted Candy?
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了POJ2479 Maximum sum[DP|最大子段和]相关的知识,希望对你有一定的参考价值。
Maximum sum
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 39599 | Accepted: 12370 |
Description
Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
Your task is to calculate d(A).
Input
The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.
Output
Print exactly one line for each test case. The line should contain the integer d(A).
Sample Input
1 10 1 -1 2 2 3 -3 4 -4 5 -5
Sample Output
13
Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.
Huge input,scanf is recommended.
Huge input,scanf is recommended.
Source
POJ Contest,Author:[email protected]
题意:求两段和最大
一开始自己想
d[i][0]前i个以i结尾选了一段
d[i][1]前i个以i结尾选了两段
然后扫描维护一个d[i][0]的最大值mx,转移
d[i][0]=max(0,d[i-1][0])+a[i];
d[i][1]=max(d[i-1][1],mx)+a[i];
初始化注意一下就行了
还有一种做法:
双向求最大字段和,最后枚举第一段的结束位置求
//两个dp函数,两种方法 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int N=5e4+5,INF=1e9; inline int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} return x*f; } int T,n,a[N]; int d[N][2],ans; void dp(){ ans=-INF;int mx=a[1]; d[1][0]=a[1];d[1][1]=-INF; for(int i=2;i<=n;i++){ d[i][0]=max(0,d[i-1][0])+a[i]; d[i][1]=max(d[i-1][1],mx)+a[i]; mx=max(mx,d[i][0]); ans=max(ans,d[i][1]); } } void dp2(){ ans=-INF; for(int i=1;i<=n;i++) d[i][0]=max(0,d[i-1][0])+a[i]; d[n+1][1]=0; for(int i=n;i>=1;i--) d[i][1]=max(0,d[i+1][1])+a[i]; int mx=d[1][0]; for(int i=2;i<=n;i++){ ans=max(ans,mx+d[i][1]); mx=max(mx,d[i][0]); } } int main(int argc, const char * argv[]) { T=read(); while(T--){ n=read(); for(int i=1;i<=n;i++) a[i]=read(); dp(); printf("%d\n",ans); } return 0; }
以上是关于POJ2479 Maximum sum[DP|最大子段和]的主要内容,如果未能解决你的问题,请参考以下文章