【动态规划】【记忆化搜索】hdu5965 扫雷
f(i,j,k)表示第i行,放的雷的状态为j{0表示不放,1表示往上放,2表示往下放,3表示上下都放},剩余还有k(0<=k<=2)个要放的方案数。
先给出我这个sb写的错误代码,死都没调出来。优越的做法在后面
#include<cstdio> #include<cstring> using namespace std; #define MOD 100000007 int T,n; char a[10010]; int f[10010][4][10]; int main(){ scanf("%d",&T); for(;T;--T){ scanf("%s",a+1); n=strlen(a+1); a[0]='0'; memset(f,0,sizeof(f)); f[1][0][a[1]-'0']=1; if(a[1]-'0'-1>=0){ f[1][1][a[1]-'0'-1]=1; } if(a[1]-'0'-1>=0){ f[1][2][a[1]-'0'-1]=1; } if(a[1]-'0'-2>=0){ f[1][3][a[1]-'0'-2]=1; } for(int i=1;i<n;++i){ for(int j=a[i]-'0';j>=a[i]-'0'-2 && j>=0;--j){ if(j<=2 && j+0<=a[i+1]-'0'){ if(j==0){ f[i+1][0][a[i+1]-'0']=(f[i+1][0][a[i+1]-'0']+f[i][0][0])%MOD; } else if(j==1){ f[i+1][1][a[i+1]-'0'-1]=(f[i+1][1][a[i+1]-'0'-1]+f[i][0][1])%MOD; f[i+1][2][a[i+1]-'0'-1]=(f[i+1][2][a[i+1]-'0'-1]+f[i][0][1])%MOD; } else{ f[i+1][3][a[i+1]-'0'-2]=(f[i+1][3][a[i+1]-'0'-2]+f[i][0][2])%MOD; } } } for(int j=a[i]-'0'-1;j>=a[i]-'0'-2-1 && j>=0;--j){ if(j<=2 && j+1<=a[i+1]-'0'){ if(j==0){ f[i+1][0][a[i+1]-'0'-1]=(f[i+1][0][a[i+1]-'0'-1]+f[i][1][0])%MOD; } else if(j==1){ f[i+1][1][a[i+1]-'0'-2]=(f[i+1][1][a[i+1]-'0'-2]+f[i][1][1])%MOD; f[i+1][2][a[i+1]-'0'-2]=(f[i+1][2][a[i+1]-'0'-2]+f[i][1][1])%MOD; } else{ f[i+1][3][a[i+1]-'0'-3]=(f[i+1][3][a[i+1]-'0'-3]+f[i][1][2])%MOD; } } } for(int j=a[i]-'0'-1;j>=a[i]-'0'-2-1 && j>=0;--j){ if(j<=2 && j+1<=a[i+1]-'0'){ if(j==0){ f[i+1][0][a[i+1]-'0'-1]=(f[i+1][0][a[i+1]-'0'-1]+f[i][2][0])%MOD; } else if(j==1){ f[i+1][1][a[i+1]-'0'-2]=(f[i+1][1][a[i+1]-'0'-2]+f[i][2][1])%MOD; f[i+1][2][a[i+1]-'0'-2]=(f[i+1][2][a[i+1]-'0'-2]+f[i][2][1])%MOD; } else{ f[i+1][3][a[i+1]-'0'-3]=(f[i+1][3][a[i+1]-'0'-3]+f[i][2][2])%MOD; } } } for(int j=a[i]-'0'-2;j>=a[i]-'0'-2-2 && j>=0;--j){ if(j<=2 && j+2<=a[i+1]-'0'){ if(j==0){ f[i+1][0][a[i+1]-'0'-2]=(f[i+1][0][a[i+1]-'0'-2]+f[i][3][0])%MOD; } else if(j==1){ f[i+1][1][a[i+1]-'0'-3]=(f[i+1][1][a[i+1]-'0'-3]+f[i][3][1])%MOD; f[i+1][2][a[i+1]-'0'-3]=(f[i+1][2][a[i+1]-'0'-3]+f[i][3][1])%MOD; } else{ f[i+1][3][a[i+1]-'0'-4]=(f[i+1][3][a[i+1]-'0'-4]+f[i][3][2])%MOD; } } } } printf("%d\n",f[n][0][0]+f[n][1][0]+f[n][2][0]+f[n][3][0]); } return 0; }
然后是斓爷优越的记忆化搜索
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> using namespace std; long long use[11000][3][3],dp[11000][3][3],qw,wq,l; char s[11000]; long long mo=100000007; long long num[3]={1,2,1}; long long getans(long long d,long long a,long long b) { if (d>l) { if (b!=0) return 0; return 1; } if (b>2 || a+b>s[d-1]-48) return 0; if (use[d][a][b]==qw) return dp[d][a][b]; use[d][a][b]=qw; dp[d][a][b]=getans(d+1,b,s[d-1]-48-a-b)*num[b]%mo; return dp[d][a][b]; } int main() { scanf("%lld",&wq); for (qw=1;qw<=wq;qw++) { scanf("%s",&s); l=strlen(s); printf("%lld\n",(getans(1,0,0)+getans(1,0,1)+getans(1,0,2))%mo); } }
——The Solution By AutSky_JadeK From UESTC
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