zojDakar Rally（01背包）

01背包 加上每次更新解题数目最多 总用时最少 因为要保证用时最少，要先把时长由小到大排序。

没排序 WA了几小时。。链接

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 #include<map>
 using namespace std;
 #define N 1010
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 int dp[N*][];
 struct node
 {
     int ti,v;
 }p[N];
 bool cmp(node x,node y)
 {
     return x.ti<y.ti;
 }
 int main()
 {
     int i,j,t,n,nn;
     cin>>nn;
     while(nn--)
     {
         memset(dp,,sizeof(dp));
         cin>>t>>n;
         for(i = ; i <= n; i++)
         cin>>p[i].ti;
         for(i = ; i <= n; i++)
         cin>>p[i].v;
         sort(p+,p+n+,cmp);
         for(i =  ; i <= n ; i++)
         {
             for(j = t ; j >= p[i].ti ; j--)
             {
                 if(dp[j-p[i].ti][]+p[i].v>dp[j][])
                 {
                     dp[j][] = dp[j-p[i].ti][]+p[i].v;
                     dp[j][] = dp[j-p[i].ti][]+;
                     dp[j][] = dp[j-p[i].ti][]+j;
                 }
                 else if(dp[j-p[i].ti][]+p[i].v==dp[j][]&&dp[j-p[i].ti][]+>=dp[j][])
                 {
                     if(dp[j-p[i].ti][]+>dp[j][]||dp[j][]>dp[j-p[i].ti][]+j)
                     {
                         dp[j][] = dp[j-p[i].ti][]+p[i].v;
                         dp[j][] = dp[j-p[i].ti][]+;
                         dp[j][] = dp[j-p[i].ti][]+j;
                     }
                 }
             }
         }
         int ans = ,x=,y=;
         for(i = t ; i >= ;i--)
         {
             if(dp[i][]>=ans)
             {
                 if(dp[i][]==ans&&dp[i][]>=x)
                 {
                     if(dp[i][]>x||dp[i][]<y)
                     {
                         ans = dp[i][];
                         x = dp[i][];
                         y = dp[i][];
                     }
                 }
                 if(dp[i][]>ans)
                 {
                     ans = dp[i][];
                     x = dp[i][];
                     y = dp[i][];
                 }
             }
         }
         cout<<ans<<" "<<x<<" "<<y<<endl;
     }
     return ;
 }