POJ 3254 Corn Fields
2017-02-22 23:53
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Description
Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares
are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice
as to which squares to plant.
Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways
he can choose the squares to plant.
Input
Line 1: Two space-separated integers: M and N
Lines 2..M+1: Line i+1 describes row i of the pasture with N space-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)
Output
Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.
Sample Input
Sample Output
Hint
Number the squares as follows:
There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9.
Source
USACO 2006 November Gold
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
状压DP~
基础题?
用二进制数记录每一行的状态(先预处理出每行中草不相邻的状态),然后通过位运算判断是否上下相邻以及符合该行的要求,最后直接把所有状态加起来就可以了~
检查了一节课啊……在学校输出样例还是8回来就是9了……为什么啊好奇怪……
#include<cstdio>
#define modd 100000000
int n,m,x,a[13],cas[8200],f[8200][13],tot,ans;
bool che(int u)
{
return !(u&(u<<1));
}
int main()
{
scanf("%d%d",&n,&m);tot=(1<<m)-1;
for(int i=1;i<=n;i++)
for(int j=0;j<m;j++) scanf("%d",&x),a[i]+=(x^1)*(1<<j);
for(int i=0;i<=tot;i++)
if(che(i))
{
cas[++cas[0]]=i;
if(!(i&a[1])) f[cas[0]][1]=1;
}
for(int i=2;i<=n;i++)
for(int j=1;j<=cas[0];j++)
if(!(cas[j]&a[i]))
{
for(int k=1;k<=cas[0];k++)
if(!(cas[k]&a[i-1]) && !(cas[k]&cas[j])) f[j][i]=(f[j][i]+f[k][i-1])%modd;
}
for(int i=1;i<=cas[0];i++) ans=(ans+f[i]
)%modd;
printf("%d\n",ans);
return 0;
}
Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares
are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice
as to which squares to plant.
Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways
he can choose the squares to plant.
Input
Line 1: Two space-separated integers: M and N
Lines 2..M+1: Line i+1 describes row i of the pasture with N space-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)
Output
Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.
Sample Input
2 3 1 1 1 0 1 0
Sample Output
9
Hint
Number the squares as follows:
1 2 3 4
There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9.
Source
USACO 2006 November Gold
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
状压DP~
基础题?
用二进制数记录每一行的状态(先预处理出每行中草不相邻的状态),然后通过位运算判断是否上下相邻以及符合该行的要求,最后直接把所有状态加起来就可以了~
检查了一节课啊……在学校输出样例还是8回来就是9了……为什么啊好奇怪……
#include<cstdio>
#define modd 100000000
int n,m,x,a[13],cas[8200],f[8200][13],tot,ans;
bool che(int u)
{
return !(u&(u<<1));
}
int main()
{
scanf("%d%d",&n,&m);tot=(1<<m)-1;
for(int i=1;i<=n;i++)
for(int j=0;j<m;j++) scanf("%d",&x),a[i]+=(x^1)*(1<<j);
for(int i=0;i<=tot;i++)
if(che(i))
{
cas[++cas[0]]=i;
if(!(i&a[1])) f[cas[0]][1]=1;
}
for(int i=2;i<=n;i++)
for(int j=1;j<=cas[0];j++)
if(!(cas[j]&a[i]))
{
for(int k=1;k<=cas[0];k++)
if(!(cas[k]&a[i-1]) && !(cas[k]&cas[j])) f[j][i]=(f[j][i]+f[k][i-1])%modd;
}
for(int i=1;i<=cas[0];i++) ans=(ans+f[i]
)%modd;
printf("%d\n",ans);
return 0;
}
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