hdu6078 wave sequence dp

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http://acm.hdu.edu.cn/showproblem.php?pid=6078

题意:题意很清楚,求满足条件的序列个数

二维dp,dp【i】【j】【k】代表以a【i】,b【j】结尾的形状为k的波浪序列的个数,形状有两种,波峰波谷,也就是要比前一个大或者小的状态

dp【i】【j】【0】=sum(dp【z】【x】【1】)(z<i,x<j&&a[z]>a[i])

dp【i】【j】【1】=sum(dp【z】【x】【0】)(z<i,x<j&&a[z]<a[i])


直接转移的复杂度比较高,利用前缀和思想,s【i】【j】【k】表示以a【z】(0<z<=i),b【j】结尾形状为k的波浪序列个数

s【i】【j】【k】=sum(dp【z】【j】【k】)(z<=i)

枚举a【z】,枚举b【x】,当遇到a【z】等于b【x】时才能转移,且转移的都是符合上述条件的,这样可以在枚举前面的b【x】时就记录下来所有比a【z】大和所有比a【z】小的元素的方案数,这样能把所有满足上面dp转移方程里x<j的都枚举到,通过前缀和又可以把所有满足z<i的也枚举到


这样会发现根本不需要dp【i】【j】【k】



#include<bits/stdc++.h>
using namespace std;
const int maxn=2005;
const int mo=998244353;
long long dp[maxn][maxn][2],s[maxn][maxn][2],a[maxn],b[maxn];
int main()
{
    int T_T,i,j,n,m;
    cin>>T_T;
    while(T_T--)
    {
        cin>>n>>m;
        for(i=1;i<=n;i++)
            cin>>a[i];
        for(i=1;i<=m;i++)
            cin>>b[i];
        memset(dp,0,sizeof(dp));
        memset(s,0,sizeof(s));
        for(i=1;i<=n;i++)
        {
            long long ans1=1,ans2=0;
            for(j=1;j<=m;j++)
            {
            	s[i][j][0]=s[i-1][j][0];
            	s[i][j][1]=s[i-1][j][1];
                if(b[j]==a[i])
                {
                    s[i][j][0]=(s[i][j][0]+ans1)%mo;
                    s[i][j][1]=(s[i][j][1]+ans2)%mo;
                }
                else if(b[j]>a[i])
                    ans1+=s[i-1][j][1];
                else
                    ans2+=s[i-1][j][0];
            }
        }
        long long ans=0;
        for(i=1;i<=m;i++)
            ans=(ans+s[n][i][0]+s[n][i][1])%mo;
        cout<<ans<<endl;
    }
    return 0;
}


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第四场多校,到现在才补……         又是类似最长公共子串类型的dp。大致题意是给你两个序列a和b,然后问你能够选出多少个f和g映射,使得a[fi]==b[fi],而且a[f1]、a[f2]、a[f3]……a[fn]是波浪序列。         很容易想到的一个递推是dp[i][j][t]=Σdp[x][y][t^1](a[i]==b[j]且a[x]==b[y],同时满足波浪性),其中dp[i][j][t]表示序列a取到第i位,序列b取到第j位,而且当前是波峰(t==1)或者波谷(t==0)时的方案数

alpc_qleonardo 765

HDU 6078 Wavel Sequence (dp)

Description Have you ever seen the wave? It’s a wonderful view of nature. Little Q is attracted to such wonderful thing, he even likes everything that looks like wave. Formally, he defines a sequence

小坏蛋_千千 763

17暑假多校联赛4.12 HDU 6078 Wavel Sequence

Wavel SequenceTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)  Problem DescriptionHave you ever seen the wave? It’s a wonderful view of nature. Little Q is attracte

小学生 450

HDU 6078 Wavel Sequence (dp + 树状数组, 2017 Multi-Univ Training Contest 4)

Problem含 N 个数的 a 数组 a1,a2,⋯,ana_1, a_2,\cdots, a_n 和含 M 个数的 b 数组 b1,b2,⋯,bmb_1, b_2,\cdots, b_m ,求有多少 a,b 公共子序列(记作 arr, 长为 len )满足 arr1<arr2>arr3<arr4>arr5⋯arr_1 \lt arr_2\gt arr_3\lt arr_4 \gt arr_5\c

DorMOUSENone的博客 798

HDU6078 Wavel Sequence(动态规划)

Wavel Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 728    Accepted Submission(s): 377 Problem Description Have you ev

say_c_box的博客 502

二分答案+划分DP:复制书稿与洛谷P1281最小化最大值

在算法与调度场景中,序列划分常要求把连续元素切成若干段,并让各段代价的最大值尽可能小,这就是典型的最小化最大值问题。其核心原理是可行性具有单调性:给定上界 x,若能完成划分,则更大上界也可行,因此可用二分答案把最优化转成判定;若约束更复杂,划分型动态规划可通过枚举最后一段完成全局最优。技术价值在于把组合搜索压缩为对数级判定,并结合贪心与从后往前还原,处理多解时“前面的人尽量少抄”的方向性要求。类似模型广泛出现在数列分段、Monthly Expense、任务调度与书稿复制中。围绕复制书稿与洛谷P1281,二分

weixin_34054866的博客 398

2017多校训练Contest4: 1012 Wavel Sequence hdu6078

Problem Description Have you ever seen the wave? It's a wonderful view of nature. Little Q is attracted to such wonderful thing, he even likes everything that looks like wave. Formally, he defines

lqybzx的专栏 459
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