hdu 1231, dp ,maximum consecutive sum of integers, find the boundaries, possibly all negative, C++ 分类: hdoj 2015-07-12 03:24 87人阅读 评论(0) 收藏

the algorithm of three version below is essentially the same, namely, Kadane’s algorithm, which is of O(n) complexity. https://en.wikipedia.org/wiki/Maximum_subarray_problem

the evolution of the implementations is to remove redundancy and do what is really needed, the side effect of doing so is becoming more efficient.
IMHO, version 1.0.2 is well commented which shows the guidelines of efficient bookkeeping of boundaries, first and last can also be easily changed to first_iter and last_iter to print the whole subarray if needed.
// version 1.0.0, a coordinate array, a traversing to find the first element

#include <cstdio>
#include <algorithm>

#define MAXSIZE 10005

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif
    int N,i,tmp,*p;
    int nums[MAXSIZE]={-1}, dp[MAXSIZE]={0};
    while(scanf("%d",&N)==1 && N>0) {
        for(i=1;i<=N;++i) { scanf("%d",&nums[i]); }
        for(i=1;i<=N;++i) {
            tmp=nums[i]+(dp[i-1]>0?dp[i-1]:0);
            dp[i]=tmp>0?tmp:0;
            //dp[i]=std:max(0,nums[i]+std::max(dp[i-1],0));
        }
        p=std::max_element(dp,dp+N+1);
        if(p==dp) {
            if(nums==std::max_element(nums,nums+N+1)) { i=1,tmp=N; }
            else {
                for(i=0;i<=N && nums[++i]<0;) {}
                for(tmp=i;nums[++tmp]==0;) {} --tmp;
            }
        }
        else {
            for(tmp=i=p-dp;i>0 && dp[--i]>0;) {}
            for(;i>0 && nums[i]==0 && dp[--i]==0;) {} ++i;
        }
        printf("%d %d %d
",*p,nums[i],nums[tmp]);
    }
    return 0;
}

// version 1.0.1, no coordinate array, modifying the data, a traversing to find first element

#include <cstdio>
#include <algorithm>

#define MAXSIZE 10005

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif
    int N,i,j,tmp,last,sum;
    int nums[MAXSIZE]={-1};
    while(scanf("%d",&N)==1 && N>0) {
        for(i=1;i<=N;++i) { scanf("%d",&nums[i]); }
        for(sum=-1,last=nums[N],j=0, i=1;i<=N;++i) {
            tmp=nums[i]+(nums[i-1]>0?nums[i-1]:0);
            if(tmp>=0) {
                if(tmp>sum) { sum=tmp; last=nums[i]; j=i; }
                nums[i]=tmp;
            }
        }
        if(sum==-1) ++sum;
        else for(;j>0 && nums[--j]>=0;) {}
        printf("%d %d %d
",sum,nums[j+1],last);
    }
    return 0;
}

// version 1.0.2, no coordinate array, not modify data, no extra traversing to find boundary element

#include <cstdio>
#include <algorithm>

#define MAXSIZE 10005

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif
    // prev -- maxsum ending here, sum -- maxsum so far, res -- result
    int N,i,first,last,tmp,sum,res,prev;
    int nums[MAXSIZE];
    while(scanf("%d",&N)==1 && N>0) {
        for(i=0;i<N;++i) { scanf("%d",&nums[i]); }
        for(res=prev=sum=-1,first=nums[0],last=nums[N-1], i=0;i<N;++i) {
            if(prev<0) {
                if(nums[i]>=0) {
                    // prev start increasing, update candidate of first -- tmp
                    tmp=prev=nums[i];
                    // update candidate of result -- sum
                    if(prev>sum) { sum=prev; }
                }
            }
            else {
                prev+=nums[i];
                // prev stop increasing, update first, last, res
                if(nums[i]<=0) { if(sum>res) { res=sum; first=tmp; last=nums[i-1]; } }
                // update candidate of result -- sum
                else if(prev>sum) { sum=prev; }
            }
        }
        // update first, last, res, -- only if partial sum remain increasing
        if(sum>res) { res=sum; first=tmp; last=nums[i-1]; }
        // all negative
        if(res==-1) ++res;
        printf("%d %d %d
",res,first,last);
    }
    return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。// p.s. If in any way improment can be achieved, better performance or whatever, it will be well-appreciated to let me know, thanks in advance.

原文地址:https://www.cnblogs.com/qeatzy/p/4716218.html