Pebbles
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 795 Accepted Submission(s): 439
The player distributes pebbles across the board so that:
?At most one pebble resides in any given square.
?No two pebbles are placed on adjacent squares. Two squares are considered adjacent if they are horizontal, vertical, or even diagonal neighbors. There's no board wrap, so 44 and 61 of row three aren't neighbors. Neither are 33 and 75 nor 55 and 92.
The goal is to maximize the number of points claimed by your placement of pebbles.
Write a program that reads in a sequence of boards from an input file and prints to stdout the maximum number of points attainable by an optimal pebble placement for each.
85 50 74 94 28
92 96 23 71 10
23 61 31 30 46
64 33 32 95 89
78 78 11 55 20 11
98 54 81 43 39 97
12 15 79 99 58 10
13 79 83 65 34 17
85 59 61 12 58 97
40 63 97 85 66 90
33 49 78 79 30 16 34 88 54 39 26
80 21 32 71 89 63 39 52 90 14 89
49 66 33 19 45 61 31 29 84 98 58
36 53 35 33 88 90 19 23 76 23 76
77 27 25 42 70 36 35 91 17 79 43
33 85 33 59 47 46 63 75 98 96 55
75 88 10 57 85 71 34 10 59 84 45
29 34 43 46 75 28 47 63 48 16 19
62 57 91 85 89 70 80 30 19 38 14
61 35 36 20 38 18 89 64 63 88 83
45 46 89 53 83 59 48 45 87 98 21
15 95 24 35 79 35 55 66 91 95 86 87
94 15 84 42 88 83 64 50 22 99 13 32
85 12 43 39 41 23 35 97 54 98 18 85
84 61 77 96 49 38 75 95 16 71 22 14
18 72 97 94 43 18 59 78 33 80 68 59
26 94 78 87 78 92 59 83 26 88 91 91
34 84 53 98 83 49 60 11 55 17 51 75
29 80 14 79 15 18 94 39 69 24 93 41
66 64 88 82 21 56 16 41 57 74 51 79
49 15 59 21 37 27 78 41 38 82 19 62
54 91 47 29 38 67 52 92 81 99 11 27
31 62 32 97 42 93 43 79 88 44 54 48
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std; int a[][];
int state[],len;//
int Num[][];
int befor[];
int now[];
void Init()
{
int i,k=<<;
len=;
for(i=;i<k;i++)
{
if( (i&(i>>)) || (i&(i<<)) );
else
state[len++]=i;
}
}
void make_init(int n)
{
int i,j,k=<<n,ans,x;
memset(Num,,sizeof(Num));
for(i=;i<=n;i++)
{
for(j=;j<len&&state[j]<k;j++)
{
x=state[j];
ans=;
while(x)
{
if(x&)
{
Num[i][j]+=a[i][ans];
}
x=x>>;
ans++;
}
}
}
}
void solve(int n)
{
int i,j,k=<<n,s,hxl;
memset(befor,,sizeof(befor));
memset(now,,sizeof(now));
for(i=;i<=n;i++)
{
for(j=;j<len&&state[j]<k;j++)
{
hxl=;
for(s=;s<len&&state[s]<k;s++)
{
if( (state[j]&state[s]) )continue;
if( ((state[j]<<)&state[s]) || ((state[j]>>)&state[s]) ) continue;
if(hxl<befor[s])
hxl=befor[s];
}
now[j]=hxl+Num[i][j];
}
for(j=;j<len&&state[j]<k;j++)
{
befor[j]=now[j];
now[j]=;
}
}
for(i=,hxl=;i<len&&state[i]<k;i++)
if(hxl<befor[i])
hxl=befor[i];
printf("%d\n",hxl);
}
int main()
{
int n;
int i,j,tom,k;
char c[];
Init();
while(gets(c+)>)
{
k=strlen(c+);
if(k==)continue;
for(tom=,i=,n=;i<=k;i++)
{
if(c[i]>=''&&c[i]<='')
{
tom=tom*+c[i]-'';
}
else
{
a[][++n]=tom;
tom=;
}
}
a[][++n]=tom; for(i=;i<=n;i++)
for(j=;j<=n;j++)
scanf("%d",&a[i][j]); make_init(n);
solve(n);
}
return ;
}