An exhilarating sale.

First, the smooth sale
"Problem description"
Mrkill abroad to buy back a lot of things, look on the annoying, really annoying.
On his way home, he suddenly received a notice, not a lot of overseas goods, this is a good reason to
Stall his friends, this pile of things can be sold to overseas cattle to make a difference. A total of n items were counted,
Each item has the weight of WI, the value of pi, but his suitcase is only a load of M, that is, he only
I can sell some of the items, and of course he wants the most money, so he can buy the most games.
Mrkill looking at this pile of products that are about to become steam wallets:
"Hey, that smells good."
"Input Format"
The input file name is sell.in.
Input data first behavior N,m next n lines are WI and pi, respectively
"Output Format"
The input file name is sell.out.
An integer representing the maximum value
"Input and output Example 1"
Sell.in Sell.out
2 2 3
1 3
2 2

4 3 10
3 10
2 7
2 8
1 1
Example 1 Explanation: Only the first product is the best
Example 2 Explanation: Only the first product is the best
"Data size and conventions"
For 20% of data n≤20,
Another 20% of the data n≤10^5, 1≤wi≤2
Another 20% of the data n≤5*10^2, 1≤pi≤20;
For 100% data n≤10^5,m≤3*10^5, 1≤wi≤3, 1≤pi≤10^10

Because of the particularity of WI, the idea is DP, we consider the three kinds of weight of items to consider separately, then the aftereffect is that each commodity in the cost of how much weight has been used several.

So it is easy to think of the value of each product in order to be able to carry out DP.

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include < Algorithm> #define LL Long long#define il inline#define db double#define max (a) > (b)? (  A):(B)) using namespace Std;il int gi () {int x=0,y=1;  Char Ch=getchar (); while (ch< ' 0 ' | |      Ch> ' 9 ') {if (ch== '-') y=-1;    Ch=getchar ();      } while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 ';    Ch=getchar (); } return x*y;}  IL ll GL () {ll x=0,y=1;  Char Ch=getchar (); while (ch< ' 0 ' | |      Ch> ' 9 ') {if (ch== '-') y=-1;    Ch=getchar ();      } while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 ';    Ch=getchar (); } return x*y;}  struct thing{ll p; int W;} t1[100045],t2[100045],t3[100045];il BOOL CMP (thing a,thing b) {return A.P&GT;B.P;}  ll F[300045][4];int main () {int n=gi (), M=gi (), Wei;  int w1=0,w2=0,w3=0;      for (int i=1;i<=n;i++) {wei=gi ();  if (wei==1) {T1[++w1].w=wei;      T1[W1].P=GL ();}  if (wei==2) {T2[++w2].w=wei; T2[W2].P=GL ();}  if (wei==3) {T3[++w3].w=wei;    T3[W3].P=GL ();}  } sort (t1+1,t1+1+w1,cmp);  Sort (t2+1,t2+1+w2,cmp);  Sort (t3+1,t3+1+w3,cmp);  for (int i=1;i<=m;i++) {if (f[i-1][0]+t1[f[i-1][1]+1].p>=f[i][0]) {F[I][0]=F[I-1][0]+T1[F[I-1][1]+1].P;  f[i][1]=f[i-1][1]+1;  F[I][2]=F[I-1][2];      F[I][3]=F[I-1][3];}  if (f[i-2][0]+t2[f[i-2][2]+1].p>=f[i][0]&&i>1) {f[i][0]=f[i-2][0]+t2[f[i-2][2]+1].p;  F[I][1]=F[I-2][1];  f[i][2]=f[i-2][2]+1;      F[I][3]=F[I-2][3];}  if (f[i-3][0]+t3[f[i-3][3]+1].p>=f[i][0]&&i>2) {f[i][0]=f[i-3][0]+t3[f[i-3][3]+1].p;  F[I][1]=F[I-3][1];  F[I][2]=F[I-3][2];    f[i][3]=f[i-3][3]+1;}  } printf ("%lld\n", f[m][0]); return 0;}

An exhilarating sale.