HDOJ 5188 zhx and contest 贪婪+01背包
zhx and contest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 455 Accepted Submission(s): 158
One day, zhx takes part in an contest. He found the contest very easy for him.
There are n problems
in the contest. He knows that he can solve the ith problem
in ti units
of time and he can get vi points.
As he is too powerful, the administrator is watching him. If he finishes the ith problem
before time li,
he will be considered to cheat.
zhx doesn't really want to solve all these boring problems. He only wants to get no less than w points.
You are supposed to tell him the minimal time he needs to spend while not being considered to cheat, or he is not able to get enough points.
Note that zhx can solve only one problem at the same time. And if he starts, he would keep working on it until it is solved. And then he submits his code in no time.
Seek EOF as
the end of the file.
For each test, there are two integers n and w separated
by a space. (1≤n≤30, 0≤w≤109)
Then come n lines which contain three integers ti,vi,li.
(1≤ti,li≤105,1≤vi≤109)
1 3
1 4 7
3 6
4 1 8
6 8 10
1 5 2
2 7
10 4 1
10 2 3
7
8
zhx is naive!
/* ***********************************************
Author :CKboss
Created Time :2015年03月19日 星期四 10时26分15秒
File Name :HDOJ5188.cpp
************************************************ */ #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map> using namespace std; const int maxn=200100; int n,w;
int sumtime,sumv;
int dp[maxn*30]; struct PB
{
int t,v,l;
}pb[50]; bool cmp(PB a,PB b)
{
return a.l-a.t<b.l-b.t;
} int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout); while(scanf("%d%d",&n,&w)!=EOF)
{
sumv=0;
for(int i=0;i<n;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
pb[i]=(PB){x,y,z};
sumv+=y;
}
if(sumv<w)
{
puts("zhx is naive!");
continue;
} sort(pb,pb+n,cmp);
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
for(int j=maxn;j>=max(pb[i].t,pb[i].l);j--)
dp[j]=max(dp[j],dp[j-pb[i].t]+pb[i].v); int ans=0;
for(int i=0;i<maxn;i++)
if(dp[i]>=w) { ans=i; break; } printf("%d\n",ans);
} return 0;
}
版权声明:来自: 代码代码猿猿AC路 http://blog.csdn.net/ck_boss
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