The Double HeLiX

最新推荐文章于 2019-04-12 08:29:33 发布

原创最新推荐文章于 2019-04-12 08:29:33 发布 · 335 阅读

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本文介绍了一个寻找两个整数序列中最大路径和的问题，并提供了一种通过处理前缀和及利用贪心策略找到最大路径和的有效算法实现。

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The Double HeLiX

Two ﬁnite, strictly increasing, integer sequences are given. Any common integer between the two sequences constitute an intersection point. Take for example the following two sequences where intersection points are
printed in bold:

First= 3 5 7 9 20 25 30 40 55 56 57 60 62
Second= 1 4 7 11 14 25 44 47 55 57 100

You can ‘walk” over these two sequences in the following way:

You may start at the beginning of any of the two sequences. Now start moving forward.
At each intersection point, you have the choice of either continuing with the same sequence you’re currently on, or switching to the other sequence.

The objective is ﬁnding a path that produces the maximum sum of data you walked over. In the above example, the largest possible sum is 450, which is the result of adding 3, 5, 7, 9, 20, 25, 44, 47, 55, 56, 57, 60, and 62

Input

Your program will be tested on a number of test cases. Each test case will be speciﬁed on two separate lines. Each line denotes a sequence and is speciﬁed using the following format:

n v1 v2 ... vn

Where n is the length of the sequence and vi is the ith element in that sequence. Each sequence will have at least one element but no more than 10,000. All elements are between -10,000 and 10,000 (inclusive).
The last line of the input includes a single zero, which is not part of the test cases.

Output

For each test case, write on a separate line, the largest possible sum that can be produced.

Sample

Input:
13 3 5 7 9 20 25 30 40 55 56 57 60 62
11 1 4 7 11 14 25 44 47 55 57 100
4 -5 100 1000 1005
3 -12 1000 1001
0

Output:
450
2100

【分许】有两个序列，当两个序列中的数字相同时，可以从序列1跳到序列2，问过程中踩过的数字和的最大值。

贪心。

先处理前缀和，然后找出相同点的位置，将每一段的的大值加上即为最大值。

#include <iostream>
#include <cstdio>
#include <stack>
#include <cstring>
using namespace std;
const int maxn = 1e4 + 10;
typedef long long LL;
#define cl(a,b) memset(a,b,sizeof a);
int a[maxn],b[maxn];
LL suma[maxn],sumb[maxn];
int visa[maxn],visb[maxn];
int main()
{
    int n,m;
    while(~scanf("%d",&n),n){
        cl(suma,0);
        cl(sumb,0);
        suma[0] = 0;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            suma[i] = suma[i-1]+a[i];
        }
        scanf("%d",&m);
        sumb[0] = 0;
        for(int i=1;i<=m;i++){
            scanf("%d",b+i);
            sumb[i] = sumb[i-1]+b[i];
        }
        int i=1,j=1,k=1;
        while(i<=n&&j<=m){
            if(a[i] < b[j]){
                i++;
            }
            else
            if(a[i] > b[j]){
                j++;
            }
            else{
                visa[k] = i;
                visb[k] = j;
                k++;
                i++;
                j++;
            }
        }
        visa[k] = n;
        visb[k] = m;
        LL ans = 0;
        for(int i=1;i<=k;i++){
            ans += max(suma[visa[i]]-suma[visa[i-1]],sumb[visb[i]]-sumb[visb[i-1]]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}