ZOJ 3723 Starfruit
恶心的状压DP。。。
与炮兵阵地类似。。。多了两个方向。而且发射的炮弹可能被石头挡住。。。。
1.因为方向是对称的，所以可以把下面两条边翻上来,考虑这样只要考虑上面的行就行了。。。。
2.被石头挡住。。。可以模拟一下
3.卡内存。。。。要用滚动数组+把可能的状态存下来
Plants VS Zombie is an interesting game, recently Edward has become totally addicted to playing this game. There are many kinds of plants he can choose to defend zombies, but what Edward interests most is a kind of plants called Starfruit. As we know, most plants stand toward right and only can attack zombies those are precisely on its right. However, starfruit is much more powerful, it can sparks little stars towards five directions!
Edward is so interested in this fruit that he wrote a game with nothing but starfruit. Unfortunately, something beyond exception happened, that starfruit now only can attack less than 3 units distance(by Euclidean distance), also he made a mistake with directions: now starfruits spark towards up, left, right, lower left and lower right(not as usual in PVZ). What's more, now starfruit attacks each other!
As Edward likes statfruit very much, he wants to put as more starfruit into the map as possible, but if you place a starfruit on a lattice attacked by another starfruit, it will die and disappear immediately. So now he wants to know the maximum number of starfruits can be placed on the map, but recently he's busy preparing Summer Camp so he turns to your help.
Input
The input consists of several cases. Each case starts with a line contains two integer n(1≤n≤1000) and m(1≤m≤12), indicates that it's a map of width m and height n. Then follows nlines each with m characters, which is either 'X' or '.'. A '.' means you can put starfruit on it, A 'X' means there is a rock there so that you can't put anything on it, also the stars sparked by starfruits can't across a rock.
Input ends when both n and m equals to 0. This case will not be executed.
Output
For each case, output the maximum number of the starfruit you can put.
Sample Input
3 3 ... XX. ... 0 0
Sample Output
3
Hint
- All lattices are squares of one unit, both starfruit and rock are considered at the center of the lattice.
- The answer to the sample: ('O' means starfruit)
O..
XXO
O.. - If a map like this and 'O' is a starfruit, we consider point A and B are been attacked.
OX.
XA.
..B
Contest: ZOJ Monthly, June 2013
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
int n,m,G[1100],hav[1100],dp[3][510][510],ZT[500],TZ[5000],Size;
char str[1100][15];
vector<int> state[1100];
set<int> vis;
bool ckleft(int p,int sta)
{
int newst=(sta<<1);
int temp=newst&(G[p]);
newst=newst&(~temp);
if(newst&sta) return false;
newst<<=1;
temp=newst&(G[p]);
newst=newst&(~temp);
if(newst&sta) return false;
return true;
}
bool ckright(int p,int sta)
{
int newst=(sta>>1);
int temp=newst&(G[p]);
newst=newst&(~temp);
if(newst&sta) return false;
newst>>=1;
temp=newst&(G[p]);
newst=newst&(~temp);
if(newst&sta) return false;
return true;
}
bool ckup(int p,int sta,int stapp,int stappp)
{
int qiang1=G[p-1],qiang2=G[p-2];
int temp=qiang1&sta;
int newst=sta&(~temp);
if(newst&stapp) return false;
temp=qiang2&newst;
newst=newst&(~temp);
if(newst&stappp) return false;
return true;
}
bool ckleftup(int p,int sta,int stapp,int stappp)
{
int qiang1=G[p-1],qiang2=G[p-2];
int newst=sta<<1;
int temp=qiang1&newst;
newst=newst&(~temp);
if(newst&stapp) return false;
newst<<=1;
temp=qiang2&newst;
newst=newst&(~temp);
if(newst&stappp) return false;
return true;
}
bool ckrightup(int p,int sta,int stapp,int stappp)
{
int qiang1=G[p-1],qiang2=G[p-2];
int newst=sta>>1;
int temp=qiang1&newst;
newst=newst&(~temp);
if(newst&stapp) return false;
newst>>=1;
temp=qiang2&newst;
newst=newst&(~temp);
if(newst&stappp) return false;
return true;
}
bool ck01(int sta0,int sta1)
{
if(sta0&sta1) return false;
if(sta0&(sta1<<1)) return false;
if(sta0&(sta1>>1)) return false;
return true;
}
int ST[3100];
int Count(int x)
{
int ans=0;
for(int i=0;i<m;i++)
{
if(x&(1<<i)) ans++;
}
return ans;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF&&n&&m)
{
memset(G,0,sizeof(G));
memset(hav,0,sizeof(hav));
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
{
scanf("%s",str[i]);
for(int j=m-1;j>=0;j--)
{
if(str[i][j]=='X')
{
G[i]|=1<<(m-1-j);
}
}
}
for(int i=0;i<n;i++)
{
state[i].clear();
for(int j=0;j<(1<<m);j++)
{
if(j&G[i]) continue;///不能种在石头上
if(ckleft(i,j)&&ckright(i,j))
{
if(ST[j]==0&&j) ST[j]=Count(j);
hav[i]++;
if(vis.count(j)==0)
{
vis.insert(j);
TZ[j]=Size;
ZT[Size++]=j;
}
state[i].push_back(TZ[j]);
}
}
}
///0...1...
int ans=0;
for(int i=0;i<hav[0];i++)
{
dp[0][state[0][i]][TZ[0]]=ST[ZT[state[0][i]]];
ans=max(ans,dp[0][state[0][i]][TZ[0]]);
}
for(int i=0;i<hav[1];i++)
{
for(int j=0;j<hav[0];j++)
{
if(ck01(ZT[state[0][j]],ZT[state[1][i]]))
{
dp[1][state[1][i]][state[0][j]]=max(dp[1][state[1][i]][state[0][j]],dp[0][state[0][j]][TZ[0]]+ST[ZT[state[1][i]]]);
ans=max(ans,dp[1][state[1][i]][state[0][j]]);
}
}
}
///2...n-1
for(int i=2;i<n;i++)
{
for(int j=0;j<hav[i];j++)
{
for(int k=0;k<hav[i-1];k++)
{
if(!ck01(ZT[state[i-1][k]],ZT[state[i][j]])) continue;
for(int l=0;l<hav[i-2];l++)
{
if(!ck01(ZT[state[i-2][l]],ZT[state[i-1][k]])) continue;
if(ckup(i,ZT[state[i][j]],ZT[state[i-1][k]],ZT[state[i-2][l]])
&&ckleftup(i,ZT[state[i][j]],ZT[state[i-1][k]],ZT[state[i-2][l]])
&&ckrightup(i,ZT[state[i][j]],ZT[state[i-1][k]],ZT[state[i-2][l]]))
{
dp[i%3][state[i][j]][state[i-1][k]]=
max(dp[i%3][state[i][j]][state[i-1][k]],dp[(i-1)%3][state[i-1][k]][state[i-2][l]]+ST[ZT[state[i][j]]]);
ans=max(ans,dp[i%3][state[i][j]][state[i-1][k]]);
}
}
}
}
}
printf("%d\n",ans);
}
return 0;
}