POJ 1185 Artillery

is the Chinese title. We all say water problem. But, good 1A it?

Title Effect:

Given the n*m matrix, when a cell has an artillery unit its upper and lower left and right sides (not including the oblique direction) is the attack range of this force. It is not possible for the two armies to attack each other. The maximum number of artillery units that can be deployed.

Problem Solving Ideas:
State compression Dp,dp[i][j][k] represents when line I is the state of J. The maximum number of artillery is arranged when the i-1 is in the K state.

Status can be pre-processed, there are only 60 kinds.

Here's the code:

#include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> #include <    Stdlib.h>using namespace Std;int min (int a,int b) {if (a>b) a=b; return A;}    int max (int a,int b) {if (a<b) a=b; return A;}    int n,m,vaild[65],cn[65],cnt;void judge () {int p,num;    cnt=0; for (int i=0; i<1<<10; i++) {if ((i<<1) &i) | |        ((i<<2) &i))        {continue;        } int temp=i;            while (temp) {if (temp%2) cn[cnt]++;        temp>>=1;    } vaild[cnt++]=i;    }}int Place[105];char s[15];int dp[105][65][65];int Main () {judge ();        while (scanf ("%d%d", &n,&m)!=eof) {int ans=0;            for (int i=1; i<=n; i++) {scanf ("%s", s);            place[i]=0;                for (int j=0; j<m; J + +) {place[i]<<=1;               if (s[j]== ' H ') {place[i]++; }}} memset (Dp,0,sizeof (DP)); for (int i=1, i<=n; i++) {if (i==1) {for (int j=0; vaild[j]<1<<m&am p;&j<cnt; J + +) {if ((Place[1]&vaild[j]) ==0) {dp[1][                    J][0]=CN[J];                } ans=max (Ans,dp[1][j][0]);  }} else if (i==2) {for (int j=0; vaild[j]<1<<m&&j<cnt; J + +) {if ((Place[2]&vaild[j]) ==0) {for (i NT K=0; vaild[k]<1<<m&&k<cnt; k++) {if ((vaild[k]&vaild[j)) ==0&& (Place[1]&vaild[k]) ==0                            ) {Dp[2][j][k]=dp[1][k][0]+cn[j]; } Ans=max (ans,dp[2][j][k]); }}}} else {for (int j=0; vaild[j]< 1<<m&&j<cnt; J + +) {if ((Place[i]&vaild[j]) ==0) {for (in T k=0; vaild[k]<1<<m&&k<cnt; k++) {if ((Vaild[k]&vaild[j]) ==0&& (place[i-1]&vaild[k]) =                                =0) {for (int l=0;vaild[l]<1<<m&&l<cnt;l++) {if ((Vaild[k]&vaild[l]) ==0&& (vaild[j]&v                                        Aild[l]) ==0&& (Place[i-2]&vaild[l]) ==0) {                                    Dp[i][j][k]=max (Dp[i][j][k],dp[i-1][k][l]+cn[j]);   }                                }                            }                         Ans=max (Ans,dp[i][j][k]);    }}}}} printf ("%d\n", ans); } return 0;}

POJ 1185 Artillery