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插头 DP code7-8
source link: http://abcdxyzk.github.io/blog/2011/03/20/alg-mdp-source4/
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插头 DP code7-8
2011-03-20 23:03:00
七、fzu_1977
#include<stdio.h>
#define N 50000 // N = all
#define M 1600000 // M = 3^m
int n,m,can[33][33],last[33][33];
int all,val[33],r[N],e[M],p[N][13],b[33],w[33],cas=0;
long long dp[2][N];
int H[N][15];
void change()
{
int i,j,tmp[33][33];
for(i=1;i<=n;i++) for(j=1;j<=m;j++) tmp[m-j+1][i] = can[i][j];
j = n; n = m; m = j;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++) can[i][j] = tmp[i][j];
}
void input()
{
int i,j;
char ch[33];
scanf("%d %d",&n,&m);
for(i=1;i<=n;i++)
{
scanf("%s",ch);
for(j=1;j<=m;j++)
if(ch[j-1] == 'O') can[i][j] = 1;
else
if(ch[j-1] == '*') can[i][j] = 2;
else
can[i][j] = 0;
}
if(n < m) change();
int k=0;
for(i=1;i<=n;i++) for(j=1;j<=m;j++)
{
if(can[i][j] == 1) k++;
last[i][j] = k;
}
}
int ok(int kk)
{
int i,l,a[15],c[15];
for(i=1;i<=m+1;i++) { a[i] = kk%3; kk /= 3; b[i] = -1; }
l = 0;
for(i=1;i<=m+1;i++)
if(a[i] == 1) c[++l] = i;
else
if(a[i] == 2)
{
if(l == 0) return 0;
b[c[l]] = i; b[i] = c[l];
l--;
}
if(l > 0) return 0;
return 1;
}
void init()
{
int i,j;
val[1] = 1;
for(i=2;i<=m+2;i++) val[i] = val[i-1]*3;
all=0;
for(i=0;i<val[m+2];i++)
{
e[i] = -1;
if(ok(i) == 1)
{
e[i] = all; r[all] = i;
for(j=1;j<=m+1;j++) p[all][j] = b[j];
all++;
}
}
for(i=0;i<all;i++)
for(j=1;j<=m+1;j++)
{
H[i][j] = r[i]/val[j]%3;
}
}
void solve()
{
int i,j,k,LL,UU,sta,u,y;
long long ans = 0;
u = 0;
for(i=0;i<all;i++) dp[u][i] = 0; dp[u][0] = 1;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
y = u; u = 1-u;
if(j == 1)
{
for(k=all-1;k>=0;k--)
{
if(e[r[k]/3] >= 0) dp[y][k] = dp[y][e[r[k]/3]];
if(r[k]%3 != 0) dp[y][k] = 0;
}
}
for(k=0;k<all;k++) dp[u][k] = 0;
for(k=0;k<all;k++)
{
LL = H[k][j]; UU = H[k][j+1];
if(can[i][j] == 0)
{
if(LL == 0 && UU == 0)
{
dp[u][k] = dp[y][k];
}
continue;
}
if(can[i][j] == 2)
{
if(LL == 0 && UU == 0)
dp[u][k] += dp[y][k];
}
if(LL == 0 && UU == 0)
{
sta = r[k] + val[j] + val[j+1]*2;
dp[u][e[sta]] += dp[y][k];
}
else
if(LL == 0)
{
dp[u][k] += dp[y][k];
sta = r[k] - UU*val[j+1] + UU*val[j];
dp[u][e[sta]] += dp[y][k];
}
else
if(UU == 0)
{
dp[u][k] += dp[y][k];
sta = r[k] - LL*val[j] + LL*val[j+1];
dp[u][e[sta]] += dp[y][k];
}
else
if(LL == 1 && UU == 1)
{
if(p[k][j+1] > 0)
{
sta = r[k]-val[j]-val[j+1]-val[p[k][j+1]];
dp[u][e[sta]] += dp[y][k];
}
}
else
if(LL == 2 && UU == 2)
{
if(p[k][j] > 0)
{
sta = r[k]-val[j]*2-val[j+1]*2+val[p[k][j]];
dp[u][e[sta]] += dp[y][k];
}
}
else
if(LL == 2 && UU == 1)
{
sta = r[k] - 2*val[j]-val[j+1];
dp[u][e[sta]] += dp[y][k];
}
else
if(LL == 1 && UU == 2)
{
if(r[k]-val[j]-val[j+1]*2 == 0 && last[i][j] == last[n][m])
{
ans += dp[y][k];
}
}
}
}
cas++;
printf("Case %d: %lld\n",cas,ans);
}
int main()
{
int i,T;
scanf("%d",&T);
m = 12;
init(); r[all] = 1000000000;
while(T-- > 0)
{
input();
for(i=0;r[i]<val[m+2];i++); all = i;
solve();
}
return 0;
}八、pku_3133
#include<stdio.h>
#define N 60000+100 // 3^(m+1)
int n,m,can[33][33];
int dp[2][N],H[N][13],val[33];
void solve()
{
int i,j,k,LL,UU,all,sta,u,y;
all = val[m+2]; u = 0;
for(i=0;i<all;i++) dp[u][i] = 1000000;
dp[u][0] = 0;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
y = u; u = 1-u;
if(j == 1)
{
for(k=all-1;k>=0;k--)
{
dp[y][k] = dp[y][k/3];
if(k%3 != 0) dp[y][k] = 1000000;
}
}
for(k=0;k<all;k++) dp[u][k] = 1000000;
for(k=0;k<all;k++)
{
LL = H[k][j]; UU = H[k][j+1];
if(can[i][j] == 1)
{
if(LL == 0 && UU == 0)
{
dp[u][k] = dp[y][k];
}
continue;
}
if(can[i][j] == 0)
{
if(LL == 0 && UU == 0)
{
if(dp[u][k] > dp[y][k]) dp[u][k] = dp[y][k];
sta = k+val[j]+val[j+1];
if(dp[u][sta] > dp[y][k] + 2) dp[u][sta] = dp[y][k]+2;
sta = k+(val[j]+val[j+1])*2;
if(dp[u][sta] > dp[y][k] + 2) dp[u][sta] = dp[y][k]+2;
}
else
if(LL == 0)
{
if(dp[u][k] > dp[y][k]+1) dp[u][k] = dp[y][k]+1;
sta = k-val[j+1]*UU+val[j]*UU;
if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
}
else
if(UU == 0)
{
if(dp[u][k] > dp[y][k]+1) dp[u][k] = dp[y][k]+1;
sta = k-val[j]*LL+val[j+1]*LL;
if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
}
else
if(LL == 1 && UU == 1)
{
sta = k-val[j]-val[j+1];
if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
}
else
if(LL == 2 && UU == 2)
{
sta = k-(val[j]+val[j+1])*2;
if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
}
}
else
if(can[i][j] == 2)
{
if(LL == 0 && UU == 0)
{
sta = k+val[j];
if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
sta = k+val[j+1];
if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
}
else
if((LL == 1 && UU == 0) || (LL == 0 && UU == 1))
{
sta = k-LL*val[j]-UU*val[j+1];
if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
}
}
else
if(can[i][j] == 3)
{
if(LL == 0 && UU == 0)
{
sta = k+val[j]*2;
if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
sta = k+val[j+1]*2;
if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
}
else
if((LL == 2 && UU == 0) || (LL == 0 && UU == 2))
{
sta = k-LL*val[j]-UU*val[j+1];
if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
}
}
}
}
if(dp[u][0] == 1000000) dp[u][0] = 0;
printf("%d\n",dp[u][0]);
}
int main()
{
int i,j;
val[1] = 1;
for(i=2;i<=9+2;i++) val[i] = val[i-1]*3;
for(i=0;i<val[9+2];i++) for(j=1;j<=9+1;j++) H[i][j] = i/val[j]%3;
while(scanf("%d %d",&n,&m) != EOF)
{
if(n == 0 && m == 0)break;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++) scanf("%d",&can[i][j]);
solve();
}
return 0;
}Recommend
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