hdu 6444 Neko's loop 单调队列优化DP

Posted 2021-01-01 draymonder

Neko‘s loop

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 356    Accepted Submission(s): 56

Problem Description

Neko has a loop of size n.
The loop has a happy value ai on the i−th(0≤i≤n−1) grid. 
Neko likes to jump on the loop.She can start at anywhere. If she stands at i−th grid, she will get ai happy value, and she can spend one unit energy to go to ((i+k)modn)−th grid. If she has already visited this grid, she can get happy value again. Neko can choose jump to next grid if she has energy or end at anywhere. 
Neko has m unit energies and she wants to achieve at least s happy value.
How much happy value does she need at least before she jumps so that she can get at least s happy value? Please note that the happy value which neko has is a non-negative number initially, but it can become negative number when jumping.

 

 

Input

The first line contains only one integer T(T≤50), which indicates the number of test cases. 
For each test case, the first line contains four integers n,s,m,k(1≤n≤104,1≤s≤1018,1≤m≤109,1≤k≤n).
The next line contains n integers, the i−th integer is ai−1(−109≤ai−1≤109)

 

 

Output

For each test case, output one line "Case #x: y", where x is the case number (starting from 1) and y is the answer.

 

 

Sample Input

2 3 10 5 2 3 2 1 5 20 6 3 2 3 2 1 5

 

 

Sample Output

Case #1: 0 Case #2: 2

 

 

Source

2018中国大学生程序设计竞赛 - 网络选拔赛

 

 

Recommend

chendu

 

刚开始看 觉得是n^2的暴力 ，然后被实验室的人说 必须O(n)过，最后在WA了四发以后，比赛最后半个小时A了这道题

n个数，最多跳m步，每次跳到（i+k）%n，然后求距 s 的最小差，大于s 计为0

很朴素的想法 枚举每个i从0到n-1 然后暴力跑循环节

假设循环节大小为len，最后 res = max(0, getRes(len)) *m/len + max(0, getRes(m%len)); getRes(x) 就是求一个循环节上 长度最大为x的最长子段和（注意是子段 不是子序列）

可以预处理O（n）求出每个循环节， 然后对每个循环节 求上面的结果 

 

 

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>

typedef long long ll;
using namespace std;

const int N = 1e4+10;

int n,m,k,cnt ;
ll s, MAX, v[N];
bool vis[N]; vector<ll> g[N];
ll que[N<<1],mx[N<<1],sta[N<<1];


ll solve(const vector<ll>&vv, int count) {
    int sz= vv.size();
    for(int i=0;i<sz;i++)
        que[i] = que[i+sz] = vv[i];
    sz = sz<<1;
    int st=0,ed=0; 
    ll res=0;
    for(int i=0;i<sz;i++) {
        if(i==0) 
            mx[i] = que[i];
        else 
            mx[i] = mx[i-1]+que[i];

        if(i < count) 
            res = max(res, mx[i]);
        
        while (st < ed && sta[st]+count < i) 
            st++;
        if(st < ed) 
            res = max(res, mx[i] - mx[sta[st]]);
        while (st < ed && mx[i] <= mx[sta[ed-1]])
            ed--;
        sta[ed++]=i;
    }
    return res;
}

ll getRes(const vector<ll>& vv,int step,ll top) {
    ll mod = step % vv.size(); ll kk = step/ vv.size();
    ll sum = 0;
    for(int i=0; i<vv.size();i++) 
        sum += vv[i];
    ll mx1 = solve(vv, mod);
    ll mx2 = solve(vv, vv.size());
    mx1 += max(0LL, sum)*kk;
    mx2 += max(0LL, sum)*((kk>2)?kk-1:0);
    return max(mx1,mx2);
}

int main ()
{
    //freopen("in.txt","r",stdin);
    int T; scanf("%d",&T);
    for(int cas=1; cas<=T; cas++) {
        memset(vis,0,sizeof(vis));
        scanf("%d %lld %d %d", &n, &s, &m, &k);
        for(int i=0;i<n;i++) 
            scanf("%lld", &v[i]);
        cnt=0; MAX=0;
        for(int i=0; i<n; i++) {
            g[cnt].clear();
            if(!vis[i]) {
                vis[i]=1;
                g[cnt].push_back(v[i]);
                for(int j=(i+k)%n; j!=i && !vis[j]; j=(j+k)%n) {
                    g[cnt].push_back(v[j]);
                    vis[j]=1;
                }
                //for(int j=0;j<g[cnt].size();j++) 
                    //cout << g[cnt][j]<<" ";
                //cout <<endl;
                MAX = max(MAX, getRes(g[cnt], m, s));    
                cnt++;    
            }
        }
        if(MAX >= s) MAX=0;
        else MAX = s-MAX;
        printf("Case #%d: %lld
", cas, MAX);
    }
    return 0;
}