HDU 1574 RP Problem __ Dynamic Programming base DP

The problem: Each thing has three attributes, moral value A, threshold B, benefit C, if a>0, less than this threshold B can do this thing, the income C is negative, conversely, if the a<0, greater than this threshold B can do, income C is positive.

Thought: Because the character value can be positive or negative, unified plus 10000 are treated as positive, the definition of dp[n] for, the character value of n when the greatest benefit, here need to mark a person if the addition of this, the product value of n has appeared, not directly updated, there is the optimal solution.

#include <fstream> #include <iostream> #include <string> #include <cstring> #include < complex> #include <math.h> #include <set> #include <vector> #include <map> #include <queue > #include <stdio.h> #include <stack> #include <algorithm> #include <list> #include <ctime > #include <memory.h> #include <ctime> #include <assert.h> #define REP (i,a,n) for (int i=a;i<n;i+ +) #define (i,a,n) for (int i=n-1;i>=a;i--) #define PB push_back #define MP Make_pair #define ALL (x) (x). Begin (), (x)
. End () #define FI #define SE second #define EPS 1e-8 #define M_PI 3.141592653589793 typedef long LL;
const LL mod=1000000007;
const int INF=0X7FFFFFFF; ll Powmod (ll A,ll b) {ll res=1;a%=mod;for (; b;b>>=1) {if (b&1) res=res*a%mod;a=a*a%mod;}
return res;}
int dp[21000],v[21000];
int a[1010],b[1010],c[1010];
using namespace Std;
    int main () {Ios::sync_with_stdio (false); int T;cin>>t;
      while (t--) {int n;cin>>n;
      for (int i=0;i<n;i++) cin>>a[i]>>b[i]>>c[i];
      Memset (Dp,0,sizeof (DP));
      memset (v,0,sizeof (v));
      V[10000]=1;
                    for (int i=0;i<n;i++) {if (a[i]>0) {for (int j=10000+b[i];j>=0;j--) {if (v[j)) {
                      if (!v[j+a[i]]) {v[j+a[i]]=1;
                    Dp[j+a[i]]=dp[j]+c[i];
                    } else{Dp[j+a[i]]=max (Dp[j+a[i]],dp[j]+c[i]);
                else{for (int j=10000+b[i];j<=20010;j++) {
                       if (V[j]) {if (!v[j+a[i]]) {v[j+a[i]]=1;
                     Dp[j+a[i]]=dp[j]+c[i];
                    } else{Dp[j+a[i]]=max (Dp[j+a[i]],dp[j]+c[i]);
           }
                }  {}} int res=0;
      for (int i=0;i<20100;i++) Res=max (Res,dp[i]);
    cout<<res<<endl;
 }
}

Winter training has been faster than a half, road long its repair far XI, I will go up and down and search!