Problem Description
One day, Resty comes to an incredible world to seek Eve -- The origin of life. Lilith, the sister of Eve, comes with him. Although Resty wants to find Eve as soon as possible, Lilith likes to play games so much that you can't make her make any move if you don't play with her.
Now they comes to the magical world and Lilish ask Resty to play with her.
The game is following :
Now the world is divided into a m * n grids by Lilith, and Lilith gives each grid a score.
So we can use a matrix to describe it.
You should come from cell(0, 0) to cell(m-1, n-1) (Up-Left to Down-Right) and try to colloct as more score as possible.
According to Lilish's rule, you can't arrive at each cell more than once.
Resty knows that Lilish will be easy to find the max score, and he doesn't want to lose the game.
So he want to find the game plan to reach the max score.
Your task is to calculate the max score that Lilish will find, the map is so small so it shouldn't be difficult for you, right?
Now they comes to the magical world and Lilish ask Resty to play with her.
The game is following :
Now the world is divided into a m * n grids by Lilith, and Lilith gives each grid a score.
So we can use a matrix to describe it.
You should come from cell(0, 0) to cell(m-1, n-1) (Up-Left to Down-Right) and try to colloct as more score as possible.
According to Lilish's rule, you can't arrive at each cell more than once.
Resty knows that Lilish will be easy to find the max score, and he doesn't want to lose the game.
So he want to find the game plan to reach the max score.
Your task is to calculate the max score that Lilish will find, the map is so small so it shouldn't be difficult for you, right?
Input
The input consists of more than one
testdata.
Process to the END OF DATA.
For each test data :
the first live give m and n. (1<=m<=8, 1<=n<=9)
following m lines, each contain n number to give you the m*n matrix.
each number in the matrix is between -2000 and 2000
Process to the END OF DATA.
For each test data :
the first live give m and n. (1<=m<=8, 1<=n<=9)
following m lines, each contain n number to give you the m*n matrix.
each number in the matrix is between -2000 and 2000
Output
Output Format is "Case ID: ANS" one line for each
data
Don't print any empty line to the output
Don't print any empty line to the output
Sample Input
2 2
1 2
3 1
3 3
0 -20 100
1 -20 -20
1 1 1
Sample Output
Case 1: 5
Case 2: 61
插头dp
左上角走到右下角，不得重复经过格子，也可以不经过，求最大分数。
两种方法，第一是上下加两行，加两列，加障碍，使题目变成简单的回路，而不是单路。
两种方法，第一是上下加两行，加两列，加障碍，使题目变成简单的回路，而不是单路。
第二在起点终点都加一个单插头处理，其他地方也应相应微调，不过我好像调得不好，一直WA……233，不过本机测了数百组数据都是对哒，所以不管了……
/*第一种方法*/ #include<queue> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int mn=977147; int i; struct na{ int x,z; na(int xx,int zz):x(xx),z(zz){} }; int n,m,x,y,z,a[21],k,en,u=0,p1,p2; bool map[21][21]; int f[2][mn+1],ans; int v[2][mn+1]; int re[21][21]; queue <na> q; inline int gx(int x,int q1,int q2){k=0;for (register int i=m+1;i;i--) k=k*3+(i==x?q1:(i==x+1?q2:a[i]));return k;} inline void up(int x,int z,int lj,bool la){ if (la) lj+=re[x/m+1][x%m+1]; x++; k=x%2; if (v[k][z]!=x) v[k][z]=x,f[k][z]=-1e9,q.push(na(x,z)); f[k][z]=max(f[k][z],lj); } int main(){ //freopen("a.in","r",stdin); register int i,j; while(scanf("%d%d",&n,&m)!=EOF){ u++; printf("Case %d: ",u); ans=-1e9; memset(map,0,sizeof(map));memset(v,0,sizeof(v));memset(re,0,sizeof(re));memset(f,0,sizeof(f)); for (i=1;i<=m;i++) for (j=1;j<=n;j++) map[i][j+2]=1; for (i=1;i<=n;i++) for (j=1;j<=m;j++) scanf("%d",&re[i+2][j]); n+=4; m+=2; en=n*m-1; for (i=1;i<=m;i++) map[i][1]=map[i][n]=1; for (i=1;i<=n;i++) map[m][i]=1; map[1][2]=1;map[m-2][n-1]=1; if (n==1&&m==1){ printf("%d ",re[1][1]); continue; } f[0][0]=0;v[0][0]=1; q.push(na(0,0)); while(!q.empty()){ na no=q.front();q.pop(); int an=f[no.x%2][no.z]; if(no.x%m==0) no.z*=3; x=no.x%m+1;y=no.x/m+1; for (i=1;i<=m+1;i++) a[i]=0; for (i=1,j=no.z;j;i++,j/=3) a[i]=j%3; if (!map[x][y])up(no.x,gx(x,0,0),an,0);else if (a[x]==1&&a[x+1]==2){ if (no.x==en) ans=max(ans,an); }else if (a[x]==2&&a[x+1]==1) up(no.x,gx(x,0,0),an,1);else if (a[x]==0&&a[x+1]==0){ if (no.x!=en&&no.x!=1) up(no.x,gx(x,0,0),an,0); if (map[x][y+1]&&map[x+1][y]) up(no.x,gx(x,1,2),an,1); }else if (a[x]==0){ if (map[x+1][y]) up(no.x,gx(x,0,a[x+1]),an,1); if (map[x][y+1]) up(no.x,gx(x,a[x+1],0),an,1); }else if (a[x+1]==0){ if (map[x+1][y]) up(no.x,gx(x,0,a[x]),an,1); if (map[x][y+1]) up(no.x,gx(x,a[x],0),an,1); }else if (a[x]==a[x+1]){ p1=p2=0; if (a[x]==1) for (j=0,i=x+2;i<=m;i++){ if (a[i]==1) j--; if (a[i]==2) j++; if (j>0&&!p1) p1=i,j--; if (j>0&&p1){p2=i;break;} }else for (j=0,i=x-1;i;i--){ if (a[i]==1) j++; if (a[i]==2) j--; if (j>0&&!p2) p2=i,j--; if (j>0&&p2){p1=i;break;} } a[p1]=1;a[p2]=2;up(no.x,gx(x,0,0),an,1); } } printf("%d ",ans); } }
/*第二种方法*/ #include<queue> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int i; struct na{ int x,z; na(int xx,int zz):x(xx),z(zz){} }; int n,m,x,y,z,a[21],k,en,u=0,p1,p2; bool map[21][21]; int f[2][1277148],ans; int v[2][1277148]; int re[21][21]; queue <na> q; inline int gx(int x,int q1,int q2){k=0;for (register int i=m+1;i;i--) k=k*3+(i==x?q1:(i==x+1?q2:a[i]));return k;} inline void up(int x,int z,int lj,bool la){ if (la) lj+=re[x/m+1][x%m+1]; x++; k=x%2; if (v[k][z]!=x) v[k][z]=x,f[k][z]=lj,q.push(na(x,z)); if (lj>f[k][z]) f[k][z]=lj; } int main(){ register int i,j; while(scanf("%d%d",&n,&m)!=EOF){ u++; ans=0; memset(map,0,sizeof(map));memset(v,0,sizeof(v));memset(f,0,sizeof(f)); en=n*m-1; for (i=1;i<=m;i++) for (j=1;j<=n;j++) map[i][j]=1; for (i=1;i<=n;i++) for (j=1;j<=m;j++) scanf("%d",&re[i][j]); if (n==1&&m==1){ printf("Case %d: %d ",u,re[1][1]); continue; } f[0][1]=0; v[0][0]=1; q.push(na(0,1)); while(!q.empty()){ na no=q.front();q.pop(); int an=f[no.x%2][no.z]; if (no.x%m==0) no.z*=3; x=no.x%m+1;y=no.x/m+1; for (i=1;i<=m+1;i++) a[i]=0; for (i=1,j=no.z;j;i++,j/=3) a[i]=j%3; if (no.x==en){ k=0; for (i=1;i<=m+1;i++) k+=a[i]!=0; if (k==1&&(a[m]==1||a[m+1]==1)&&an+re[n][m]>ans) ans=an+re[n][m]; continue; } if (a[x]==2&&a[x+1]==1){ up(no.x,gx(x,0,0),an,1); }else if (a[x]==0&&a[x+1]==0){ up(no.x,gx(x,0,0),an,0); if (map[x][y+1]&&map[x+1][y]) up(no.x,gx(x,1,2),an,1); }else if (a[x]==0){ if (map[x+1][y]) up(no.x,gx(x,0,a[x+1]),an,1); if (map[x][y+1]) up(no.x,gx(x,a[x+1],0),an,1); }else if (a[x+1]==0){ if (map[x+1][y]) up(no.x,gx(x,0,a[x]),an,1); if (map[x][y+1]) up(no.x,gx(x,a[x],0),an,1); }else if (a[x]==1&&a[x+1]==1){ p1=p2=0; for (j=0,i=x+2;i<=m+1;i++){ if (a[i]==1) j--; if (a[i]==2) j++; if (j>0&&!p1) p1=i,j--; if (j>0&&p1){p2=i;break;} } a[p1]=1;a[p2]=2; up(no.x,gx(x,0,0),an,1); }else if (a[x]==2&&a[x+1]==2){ p1=p2=0; for (j=0,i=x-1;i;i--){ if (a[i]==1) j++; if (a[i]==2) j--; if (j>0&&!p2) p2=i,j--; if (j>0&&p2){p1=i;break;} } a[p1]=1;a[p2]=2;up(no.x,gx(x,0,0),an,1); } } printf("Case %d: %d ",u,ans); } }