2012 #5 Gold miner

Gold miner

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1889 Accepted Submission(s): 740
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4341

[align=left]Problem Description[/align]
Homelesser likes playing Gold miners in class. He has to pay much attention to the teacher to avoid being noticed. So he always lose the game. After losing many times, he wants your help.

#include <stdio.h>
#include <string.h>
#include <cmath>
#include <algorithm>
using namespace std;

struct Node
{
int x;
int y;
int t;
int v;
}a[205];

bool cmp(Node pp,Node qq)
{
double px,py,qx,qy;
px=(double)pp.x,py=(double)pp.y;
qx=(double)qq.x,qy=(double)qq.y;
if(fabs(atan2(px,py)-atan2(qx,qy))>(1e-8))
{
return atan2(px,py)<atan2(qx,qy);
}
else
{
return (px*px+py*py)<(qx*qx+qy*qy);
}
}

bool compare(Node pp,Node qq)
{
double px,py,qx,qy;
px=(double)pp.x,py=(double)pp.y;
qx=(double)qq.x,qy=(double)qq.y;
if(fabs(atan2(px,py)-atan2(qx,qy))<=(1e-8))
return true;
else
return false;
}

int dp[205][40005],coc[205][40005];

int main()
{
int n,T,cas=1;
int i,j,k;
int b[205];
while(scanf("%d %d",&n,&T)!=EOF)
{
memset(b,0,sizeof(b));
for(i=1;i<=n;i++)
scanf("%d %d %d %d",&a[i].x,&a[i].y,&a[i].t,&a[i].v);
sort(a+1,a+n+1,cmp);
for(i=1;i<n;i++)
{
for(j=i+1;j<=n;j++)
{
if(compare(a[i],a[j]))
b[i]++;
else
break;
}
}
for(i=0;i<=n;i++)
{
for(j=0;j<=T;j++)
{
dp[i][j]=0;
coc[i][j]=0;
}
}

for(i=1;i<=n;i++)
{
for(j=0;j<=T;j++)
{
dp[i][j]=coc[i][j];
}

for(j=0;j+a[i].t<=T;j++)
{
dp[i][j+a[i].t]=max(dp[i][j+a[i].t],coc[i][j]+a[i].v);
if(b[i]>0)
{
coc[i+1][j+a[i].t]=max(coc[i+1][j+a[i].t],coc[i][j]+a[i].v);
}
}

for(j=0;j<=T;j++)
{
dp[i][j]=max(dp[i-1][j],dp[i][j]);
coc[i+b[i]+1][j]=max(coc[i+b[i]+1][j],dp[i][j]);
}
}

/*for(i=1;i<=n;i++)
printf("%d ",a[i].v);
printf("\n\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=T;j++)
{
printf("%d ",coc[i][j]);
}
printf("\n");
}
printf("\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=T;j++)
{
printf("%d ",dp[i][j]);
}
printf("\n");
}
printf("\n");*/

int ans=0;
for(j=0;j<=T;j++)
if(dp
[j]>ans)
ans=dp
[j];
printf("Case %d: %d\n",cas++,ans);
}
return 0;
}

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