[BZOJ4247] 挂饰

来源：互联网发布：淘宝店商品下架上架编辑：程序博客网时间：2024/05/22 14:26

传送门

http://www.lydsy.com/JudgeOnline/problem.php?id=4247

题目大意

给定n个物品，每个物品都消耗1体积，但会增加ai的空间和bi的收益，初始有1体积的空间，求最大收益

题解

dp[i,j]=max{dp[i−1,j],dp[i−1,j−a[i]+1]+b[i]}
注意处置

const    maxn=2005;var    dp:array[0..maxn,0..2*maxn]of longint;    x,y:array[0..maxn]of longint;    i,j,k:longint;    n,m,ans:longint;function max(a,b:longint):longint;begin if a>b then exit(a) else exit(b); end;procedure sort(l,r:longint);var a,b,i,j:longint;begin    i:=l; j:=r; a:=x[(l+r)div 2];    repeat        while x[i]>a do inc(i);        while a>x[j] do dec(j);        if not(i>j) then        begin            b:=x[i]; x[i]:=x[j]; x[j]:=b;            b:=y[i]; y[i]:=y[j]; y[j]:=b;            inc(i); dec(j);        end;    until i>j;    if l<j then sort(l,j);    if i<r then sort(i,r);end;begin    readln(n);    for i:=1 to n do        readln(x[i],y[i]);    sort(1,n); {x[i]}    for i:=0 to n do        for j:=0 to n+1 do            dp[i,j]:=-1000000000;    dp[0,1]:=0;    for i:=1 to n do        for j:=0 to n do            dp[i,j]:=max(dp[i-1,j],dp[i-1,max(0,j-x[i])+1]+y[i]);       ans:=0;    for i:=0 to n do        ans:=max(ans,dp[n,i]);    writeln(ans);end.

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