Soj 3664 chess towers [Full bag deformation]

Soj 3664 chess towers

This is a question for the training team last night.

The question of usaco cows.

In a two-hour competition, question a is a big question, because the meaning of the question is wrong,

A waste of nearly an hour. Read questions.

So that I am not in the mood to write this question. After returning to the dormitory, decisive a drop.

Given some different chess, each height is known, the value is known, and the number is infinite.

Find the maximum value that t can achieve when stacked together.

(Without the following conditions, this part is a bare full backpack)

Note that if a chess with a height greater than or equal to k appears, all the following chess Heights will be compressed

The original 4/5 .. Note that each chess can only be compressed once.

 

The procedure is also simple:

Perform a full backpack once and obtain 1 to 2 * t (note that the size is slightly larger here, and the compression below may be within T ).

The greatest value in height. Here we can find the largest value of 1 to T ans

Then enumerate (if any) those chess whose height is> = K and put them on the top,

It will get a compressed height and the new value of this height. And update the answer ans.

 

# Include <iostream> <br/> # include <algorithm> <br/> # include <cstring> <br/> # include <cstdio> <br/> # include <queue> <br/> # include <vector> <br/> using namespace STD; <br/> const int max = 11000; <br/> const int INF = 0x1fffffff; </P> <p> int N, T, K, DP [Max]; <br/> struct node <br/> {<br/> int V, H; <br/>} nod [Max]; <br/> bool CMP (node, node B) <br/>{< br/> return. h <B. h; <br/>}< br/> void Init () <br/>{< br/> int I; <br/> DP [0] = 0; <br/> for (I = 1; I <= 2 * t; I ++) DP [I] =-1; <br/>}< br/> int main () <br/>{< br/> int I, j, Hei, Val, K; <br/> while (scanf ("% d", & N, & T, & K) = 3) <br/> {<br/> for (I = 1; I <= N; I ++) scanf ("% d", & nod [I]. v, & nod [I]. h); <br/> sort (& nod [1], & nod [n + 1], CMP); </P> <p> Init (); </P> <p> K =-1; <br/> for (I = 1; I <= N; I ++) <br/> {<br/> If (nod [I]. h >=k) {k = I; break ;}< br/> Hei = nod [I]. h; <br/> val = nod [I]. v; <br/> for (j = 0; j <= 2 * t; j ++) <br/>{< br/> If (DP [J] =-1) continue; <br/> DP [J + Hei] = max (DP [J + Hei], DP [J] + val ); <br/>}< br/> K = I; <br/> int ans = 0; <br/> for (I = 1; I <= T; I ++) ans = max (ANS, DP [I]); <br/> If (k =-1) {printf ("% d/N", ANS); Continue ;}</P> <p> for (I = K; I <= N; I ++) <br/> {<br/> Hei = nod [I]. h; <br/> val = nod [I]. v; <br/> for (j = 0; j <= 2 * t; j ++) <br/>{< br/> If (DP [J] =-1) continue; <br/> DP [J + Hei] = max (DP [J + Hei], DP [J] + val ); <br/>}</P> <p> for (I = K; I <= N; I ++) <br/> {<br/> for (j = 0; j <= 2 * t; j ++) <br/> {<br/> If (DP [J] =-1) continue; <br/> int HH = nod [I]. H + J * 4/5; <br/> If (HH> T) break; <br/> ans = max (ANS, DP [J] + nod [I]. v); <br/>}< br/> printf ("% d/N", ANS ); <br/>}< br/> // system ("pause"); <br/> return 0; <br/>}</P> <p>