Comet OJ - Contest #11 B usiness

Topic ideas:

Obviously dp problem is the number of days as a stage, and then set inside a backpack full, as it will get the node at the end of every day, so the number of days in the last cycle but also add a loop to add nodes obtained at the end of each day.

dp [u] represents now finally able to get the number of nodes when a node u, there are several places to note, first of all there is the current number of nodes from the 2000 cycle , as w [i] and the range of k is 1000, If you say that currently there are 1000 nodes, the nodes also received 1000, so there may be up to 2000 nodes (node k get is greater than 0). Followed by the final plus to take a maximum number of nodes you get every day. Finally, remember to add the current answer some nodes.

 

Code:

 

#include<bits/stdc++.h>
using namespace std;
int dp[10010],w[10010],n,m,k,ans;
pair<int,int>p[101];//代表题目中的ai,bi
int main()
{
    memset(dp,128,sizeof(dp));
    scanf("%d%d%d",&n,&m,&k);
    for(int i = 0;i<=k;i++)
        scanf("%d",&w[i]);
    for(int i = 1;i<=m;i++)
        scanf("%d%d",&p[i].first,&p[i].second);
    dp[0] = 0;
    for(int i = 1;i<=n;i++)//天数
    {
        for(int j = 1;j<=m;j++)//存储方式
            for(int u = 2000;u>=0;u--)//当前有的节点数
                dp[u] = max(dp[u],dp[u+p[j].first]+P [J] .second);
         for ( int U = 2000 ; U> = 0 ; u - the) // plus the number of nodes obtained at the end of the day 
            DP [W U + [U]] = max (DP [U + W [U]], DP [U]); 
    } 
    for ( int I = 0 ; I <= 2000 ; I ++ ) 
        ANS = max (ANS, DP [I] + I); // node plus the final answer equal to the last obtained some of the current node on 
    the printf ( " % D " , ANS);
     return  0 ;   
}