Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 29502 | Accepted: 11424 |
Description
如果在地图中的灰色所标识的平原上部署一支炮兵部队,则图中的黑色的网格表示它能够攻击到的区域:沿横向左右各两格,沿纵向上下各两格。图上其它白色网格均攻击不到。从图上可见炮兵的攻击范围不受地形的影响。
现在,将军们规划如何部署炮兵部队,在防止误伤的前提下(保证任何两支炮兵部队之间不能互相攻击,即任何一支炮兵部队都不在其他支炮兵部队的攻击范围内),在整个地图区域内最多能够摆放多少我军的炮兵部队。
Input
接下来的N行,每一行含有连续的M个字符('P'或者'H'),中间没有空格。按顺序表示地图中每一行的数据。N <= 100;M <= 10。
Output
Sample Input
5 4
PHPP
PPHH
PPPP
PHPP
PHHP
Sample Output
6
Source
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状压dp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define lbt(x) (x & -x)
#define max(a,b) ((a) > (b) ? (a) : (b))
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define fo(i,x,y) for (int i = (x); i <= (y); i++)
#define Redge(u) for (int k = head[u]; k != -1; k = edge[k].next)
using namespace std;
const int maxn = 105,maxm = 70,INF = 1000000000;
int n,m,f[maxn][maxm][maxm],S[maxn][maxm],num[maxn][maxm];
int main()
{
int p,maxv,cnt;
cin>>n>>m;
maxv = (1 << m) - 1;
REP(i,n){
char c = getchar();
while (c != 'P' && c != 'H') c = getchar();
while (c == 'P' || c == 'H'){
p = (p << 1) + (c == 'P');
c = getchar();
}
for (int s = 0; s <= maxv; s++){
if ((s | p) != p) continue;
bool flag = true;
int t = s,tot = 0; cnt = INF;
while (t){
cnt++;
if ((t & 1) && cnt <= 2) {flag = false; break;}
else if (t & 1) cnt = 0,tot++;
t >>= 1;
}
if (flag) S[i][++S[i][0]] = s,num[i][S[i][0]] = tot;
}
}
//REP(i,n) cout<<S[i][0]<<endl;
S[0][0] = 1;
for (int i = 1; i <= S[1][0]; i++){
f[1][i][1] = num[1][i];
}
for (int i = 2; i <= n; i++){
for (int j = 1; j <= S[i][0]; j++){
for (int k = 1; k <= S[i - 1][0]; k++){
if (S[i][j] & S[i - 1][k]) continue;
for (int l = 1; l <= S[i - 2][0]; l++){
if ((S[i - 1][k] & S[i - 2][l]) || (S[i][j] & S[i - 2][l])) continue;
f[i][j][k] = max(f[i][j][k],f[i - 1][k][l] + num[i][j]);
}
}
}
}
int ans = 0;
for (int i = 1; i <= S[n][0]; i++)
for (int j = 1; j <= S[n - 1][0]; j++)
ans = max(ans,f[n][i][j]);
cout<<ans<<endl;
return 0;
}