HDU 3466 Proud Merchants

Proud merchantstime limit:1000msmemory limit:65536kbthis problem would be judged onHDU. Original id:3466
64-bit integer IO format: %i64d Java class name: Main Recently, ISea went to an ancient country. For such a long time, it is the most wealthy and powerful kingdom in the world. As a result, the people in this country is still very proud even if their nation hasn ' t been so wealthy any more.
The merchants were the most typical, each of the them only sold exactly one item, the price is Pi, but they would refuse to M Ake a trade with the If your money were less than Qi, and ISea evaluated every item a value Vi.
If he had M units of money, what ' s the maximum value ISea could get?

InputThere is several test cases in the input.

Each test case begin with the integers N, M (1≤n≤500, 1≤m≤5000), indicating the items ' number and the initial mone Y.
Then N lines follow, each line contains three numbers Pi, Qi and Vi (1≤pi≤qi≤100, 1≤vi≤1000), their meaning is I n the description.

The input terminates by end of file marker.

OutputFor each test case, output one integer, indicating maximum value ISea could get.

Sample Input

2 1010 15 105 10 53 105 10 53 5 62 7 3

Sample Output

511

SourceACM-ICPC multi-university Training Contest (3)--host by WHU: This is what Chen Guolin's csdn blog says: for (I=1; i<=n; i++)
for (j=m; j>=q[i]; j--)
Dp[j]=max (Dp[j],dp[j-p[i]]+v[i]);
To ensure that the DP equation does not have any validity j-p[i] must be compared to J first, then when the calculation I, the minimum can be counted to q[i]-p[i], so that the latter can be used to the front of the state, so the order of q[i]-p[i] can be guaranteed no effect. This is a good idea, indeed!

1#include <bits/stdc++.h>2 using namespacestd;3 Const intMAXN =505;4 intn,m,dp[5010];5 structnode{6     intp,q,v;7     BOOL operator< (ConstNode &t)Const{8         returnQ-p < T.Q-T.P;9     }Ten }D[MAXN]; One intMain () { A      while(~SCANF ("%d%d",&n,&M)) { -Memset (DP,0,sizeofDP); -          for(inti =0; i < N; ++i) thescanf" %d%d%d", &d[i]. P,&d[i]. q,&D[i]. V); -Sort (d,d+N); -          for(inti =0; i < N; ++i) { -              for(intj = M; J >= D[i]. Q; --j) +DP[J] = max (Dp[j],dp[j-d[i]. P] +D[i]. V); -         } +printf"%d\n", Dp[m]); A     } at     return 0; - } - /* - 3 Ten - 5 5 - 2 7 3 in 3 5 6 - */

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HDU 3466 Proud Merchants