POJ 2479 Maximum sum(双向DP)
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Maximum sum
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 36100 | Accepted: 11213 |
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.
1000ms。50000个数,所以每次处理的时间复杂度不能超过nlogn,否则会超时。所以要让最后扫描一次就能求出答案。
基本思路就是第一次遍历先定义2个数组,分别记录前i项和(含i)与后n-i+1项和(含i)。第二次遍历再定义2个数组。分别记录以i为终点(含i)的最大子段和与以i为起点(含i)的最大子段和。
第三次遍历再定义2个数组,分别记录第i项(含i)的之前的最大子段和与第i项(含i)的之后的最大子段和。最后遍历一遍数组求出i之前(含i)子段和与i之后(不含i)子段和的最大值就可以。
#include<stack> #include<queue> #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #pragma commment(linker,"/STACK: 102400000 102400000") #define lson a,b,l,mid,cur<<1 #define rson a,b,mid+1,r,cur<<1|1 using namespace std; const double eps=1e-6; const int MAXN=50050; int num[MAXN],n,prev[MAXN],afte[MAXN],ans1[MAXN],ans2[MAXN],fans1[MAXN],fans2[MAXN],sum; int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif // ONLINE_JUDGE int tcase; scanf("%d",&tcase); while(tcase--) { scanf("%d",&n); memset(prev,0,sizeof(prev));//前i项和(含i) memset(afte,0,sizeof(afte));//后n-i+1项和(含i) memset(ans1,0,sizeof(ans1));//以i为终点(含i)的最大子段和 memset(ans2,0,sizeof(ans2));//以i为起点(含i)的最大子段和 memset(fans1,0,sizeof(fans1));//第i项(含i)的之前的最大子段和 memset(fans2,0,sizeof(fans2));//第i项(含i)的之后的最大子段和 sum=0; for(int i=1;i<=n;i++) { scanf("%d",&num[i]); prev[i]=prev[i-1]+num[i]; sum+=num[i]; } if(n==2) { printf("%d\n",sum); continue; } for(int i=n;i>=1;i--) afte[i]=afte[i+1]+num[i]; int minn=0; for(int i=0;i<n;i++) { minn=min(prev[i],minn); ans1[i+1]=prev[i+1]-minn; //printf("%d\n",ans1[i+1]); } minn=0; for(int i=n+1;i>0;i--) { minn=min(afte[i],minn); ans2[i-1]=afte[i-1]-minn; //printf("%d\n",ans2[i-1]); } int maxx=-99999999; for(int i=1;i<=n;i++) { maxx=max(maxx,ans1[i]); fans1[i]=maxx; } maxx=-99999999; for(int i=n;i>=1;i--) { maxx=max(maxx,ans2[i]); fans2[i]=maxx; } int ans=-99999999; for(int i=1;i<n;i++) ans=max(ans,fans1[i]+fans2[i+1]);//题目规定区间不能有交集 printf("%d\n",ans); } return 0; }
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