POJ1185: Artillery (reduction of the country)

Description

Ordered the general to n*m the grid. Their artillery units were deployed on the map. A n*m the map consists of n rows of M-columns, each of which can be a mountain (denoted by "H") or plain (denoted by "P"), for example. A maximum of one artillery unit can be arranged on each Goping terrain (no artillery units can be deployed on the mountain); an artillery unit is seen in the black area of the attack on the map:

Assuming that an artillery unit is deployed on a plain that is identified in gray on the map, the black grid in the figure represents the area to which it can attack: two grids along the landscape, two in each direction. Other white meshes on the graph are not attacked.

It can be seen from the diagram that the artillery's attack range is unaffected by the topography.
Now, the generals plan how to deploy artillery troops, in the premise of preventing accidental injury (to ensure that no two artillery units can attack each other, that is, no matter what one of the artillery units in the attack range of other artillery units), in the entire map area, the maximum number of troops to be placed in the Army's artillery units.

Input

The first line consists of two positive integers separated by a space, representing N and M respectively;
For the next n rows, each row contains a continuous m-character (' P ' or ' H '). There are no spaces in the middle. Represents the data for each row in the map sequentially. N <= 100;m <= 10.

Output

Only one row, including an integer k. Indicates the maximum number of artillery units that can be placed.

Sample Input

5 4PHPPPPHHPPPPPHPPPHHP

Sample Output

6

Basically the same topic as HDU4539, just a little bit different in the inference situation.

#include <stdio.h> #include <string.h> #include <algorithm>using namespace Std;int N,m;char map[105][ 15];int mat[105][15];int now[1<<10],pre[1<<10],prepre[1<<10];int l_now,l_pre,l_prepre;int dp[1 <<10][1<<10],tem[1<<10][1<<10],tt[1<<10];void dfs (int id,int k,int p,int sum) {if (k        &GT;=M) {Now[++l_now] = p;        Tt[l_now] = sum;    Return } if (k>=2 && mat[id][k] &&! ( p& (1<< (k-2))) &&!        (p& (1<< (k-1)))) DFS (id,k+1,p| (    1<<k), sum+1); else if (k==1 && mat[id][k] &&! ( p& (1<< (k-1)))) Dfs (id,k+1,p| (    1<<k), sum+1); else if (k = = 0 && mat[id][k]) DFS (id,k+1,p| (    1<<k), sum+1); DFS (id,k+1,p,sum);}    void Solve () {int i,j,k,t;    TEM[1][1] = pre[1] = prepre[1] = 0;    L_pre = L_prepre = 1;        for (k = 0; k<n; k++) {l_now = 0;        DFS (k,0,0,0); for (i = 1; i<=l_now; i++)            for (j = 1; j<=l_pre; j + +) Dp[i][j] = 0;                for (i = 1; i<=l_now; i++) for (j = 1; j<=l_pre; j + +) for (t = 1; t<=l_prepre; t++)                    {if (Now[i] & prepre[t]) continue;                    if (Now[i] & Pre[j]) continue;                    if (Pre[j] & prepre[t]) continue;                DP[I][J] = max (dp[i][j],tem[j][t]+tt[i]);        } for (i = 1; i<=l_now; i++) for (j = 1; j<=l_pre; j + +) Tem[i][j] = Dp[i][j];        for (i = 1; i<=l_pre; i++) prepre[i] = Pre[i];        for (i = 1; i<=l_now; i++) pre[i] = Now[i];        L_prepre = L_pre;    L_pre = L_now;    }}int Main () {int i,j,ans;            while (~SCANF ("%d%d", &n,&m)) {for (i = 0; i<n; i++) {scanf ("%s", Map[i]); for (j = 0; j<m; j + +) {if (map[i][j] = = ' P ') matI                [j] = 1;            else mat[i][j] = 0;        }} solve ();        Ans = 0;        for (i = 1; i<=l_pre; i++) for (j = 1; j<=l_prepre; j + +) ans = max (ans,tem[i][j]);    printf ("%d\n", ans); } return 0;}

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POJ1185: Artillery (reduction of the country)