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Mean:
略(中文題).
analyse:
比賽中最先想到的是三維dp,但思考後發現可以壓縮為二維,狀态轉移方程:
dp[i][j]=min(dp[i][j],dp[i][j-(right+fault)]+right)
其中dp[i][j]表示:
到通過第i關為止,在總共隻有j次答題機會的情況下,總共至少需要答對dp[i][j]次
最終answer為dp[n][m].
Time complexity: O(N*M*M)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* Author: crazyacking
* Date : 2016-03-30-21.00
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
const int N=1010;
LL a[N],dp[N][N];
int main()
{
int Cas;
cin>>Cas;
while(Cas--)
{
int n,m,s,t;
cin>>n>>m>>s>>t;
for(int i=1;i<=n;++i)
{
cin>>a[i];
fill(dp[i],dp[i]+N,INT_MAX);
}
fill(dp[0],dp[0]+N,0);
int max_right=a[i]/s;
if(a[i]%s) ++max_right;
for(int right=0;right<=max_right;++right)
{
int dif=a[i]-s*right,fault=0;
if(dif>0)
{
if(dif%t) fault=dif/t+1;
else fault=dif/t;
}
fault=max(fault,0);
for(int chance=m;chance>=right+fault;--chance)
dp[i][chance]=min(dp[i][chance],dp[i-1][chance-(right+fault)]+right);
}
if(dp[n][m]<INT_MAX)
cout<<dp[n][m]<<endl;
else
cout<<"No"<<endl;
}
return 0;
}
/*
/**< 帶注釋版 */
for(int i=1;i<=n;++i) // 到第i關為止
//對于本關,答對max_right題時,不算錯題的分值也能通關
for(int right=0;right<=max_right;++right) // 枚舉答對的題數
//答對right題時,需要答錯多少題
// dp[i][j] ------到本關為止,若隻有chance次答題機會,本關答對了right題,若本關能夠通關,到本關為止總共最少需要答對多少題
// chance-(right+fault) ----- 總共答題機會為chance,本關用了right+fault次,前面所有關隻有chance-(right+fault)次機會