cerc2014 Vocabulary

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给你三个字符串,这些字符串有些单词模糊不可认了,用"?"来代表。
现在你可以用任意英文小写字母来代表它们。要求是使得给定的三个字符串中
所有的"?"被你认定的字母代替后,各不相同且按字典序出现。问有多少种方式。

Input

先给出一个数字N,代表数据组数。
接下来3*N行,每行给出一个字符串。长度<=1000 000

Output

输出结果 Mod 10^9+9

Sample Input

3
?heoret?cal
c?mputer
?cience
jagiellonia
?niversity
kra?ow
?
b
c

Sample Output

42562
52
1

http://blog.csdn.net/u012647218/article/details/42008379   不用dp果断爆啊。。。
附:之前写的RE版本
#include<iostream>#include<algorithm>#include<string>#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};#include<set>//int gcd(int a,int b){return b?gcd(b,a%b):a;}#include<vector>#include<cmath>#include<queue>#include<string.h>#include<stdlib.h>#include<cstdio>#define mod 1e9+7#define ll long longusing namespace std;int com(string a,string b){for(int i=0;i<min(a.length(),b.length());++i){if(a[i]!=b[i])return (a[i]>b[i]);}return (a.length()>b.length());}map<string,int> r1;string h;int ss;set<string> w1,w2;void dfs(int k,string x){while(k<x.length()&&x[k]!='a'-1)k++;if(k>=x.length()){w1.insert(x);return;}for(int i=0;i<26;++i){x[k]='a'+i;dfs(k+1,x);}}void dfs2(int k,string x){while(k<x.length()&&x[k]!='a'-1)k++;if(k>=x.length()&&com(x,h)){w2.insert(x);r1[x]++;return;}for(int i=0;i<26;++i){x[k]='a'+i;if(com(x,h))dfs2(k+1,x);}}void dfs3(int k,string x){while(k<x.length()&&x[k]!='a'-1)k++;if(k>=x.length()&&com(x,h)){ss+=r1[h];return;}for(int i=0;i<26;++i){x[k]='a'+i;if(com(x,h))dfs3(k+1,x);}}int main(){int t;scanf("%d",&t);while(t--){ss=0;r1.clear();w1.clear();w2.clear();string x,y,z;cin>>x>>y>>z;for(int i=0;i<x.length();++i)if(x[i]=='?')x[i]='a'-1;for(int i=0;i<y.length();++i)if(y[i]=='?')y[i]='a'-1;for(int i=0;i<z.length();++i)if(z[i]=='?')z[i]='a'-1;dfs(0,x);for(set<string>::iterator it=w1.begin();it!=w1.end();++it){h=(*it);dfs2(0,y);}for(set<string>::iterator it=w2.begin();it!=w2.end();++it){h=(*it);dfs3(0,z);}cout<<ss<<endl;}    return 0;}
下面是超时代码(复杂度10^6*27*27*27)。。大神的打表代码(复杂度4*4*27*27*27)真的很神奇。。
#include<iostream>  #include<algorithm>  #include<string>  #include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};  #include<set>//int gcd(int a,int b){return b?gcd(b,a%b):a;}  #include<vector>  #include<cmath>  #include<queue>  #include<string.h>  #include<stdlib.h>  #include<cstdio>  #define bb 1000000009#define ll long long  using namespace std;  int dp[1000005][4];int mod(int a,int b){return (a+b)%bb;}int main(){int t;cin>>t;while(t--){memset(dp,0,sizeof(dp));string x,y,z;cin>>x>>y>>z;int a=x.length(),b=y.length(),c=z.length();int p=max(max(a,b),c);for(int i=0;i<p-a;++i)x+='a'-1;for(int i=0;i<p-b;++i)y+='a'-1;for(int i=0;i<p-c;++i)z+='a'-1;dp[0][0]=1;for(int i=0;i<p;++i){int a1=x[i],b1=x[i];int a2=y[i],b2=y[i];int a3=z[i],b3=z[i];if(x[i]=='?')a1='a',b1='z';if(y[i]=='?')a2='a',b2='z';if(z[i]=='?')a3='a',b3='z';for(int j=a1;j<=b1;++j){for(int k=a2;k<=b2;++k){for(int l=a3;l<=b3;++l){if(j==k&&k==l){  // 0dp[i+1][0]=mod(dp[i+1][0],dp[i][0]);dp[i+1][1]=mod(dp[i+1][1],dp[i][1]);dp[i+1][2]=mod(dp[i+1][2],dp[i][2]);dp[i+1][3]=mod(dp[i+1][3],dp[i][3]);}else if(j==k&&k<l){  // 1dp[i+1][1]=mod(mod(dp[i+1][1],dp[i][0]),dp[i][1]);dp[i+1][3]=mod(mod(dp[i+1][3],dp[i][2]),dp[i][3]);}else if(j<k&&k==l){   // 2dp[i+1][2]=mod(mod(dp[i+1][2],dp[i][0]),dp[i][2]);dp[i+1][3]=mod(mod(dp[i+1][3],dp[i][1]),dp[i][3]);}else if(j<k&&k<l){   // 3dp[i+1][3]=mod(dp[i+1][3],dp[i][0]);dp[i+1][3]=mod(dp[i+1][3],dp[i][1]);dp[i+1][3]=mod(dp[i+1][3],dp[i][2]);dp[i+1][3]=mod(dp[i+1][3],dp[i][3]);}else if(j<k){  // ab ac badp[i+1][3]=mod(dp[i+1][3],dp[i][1]);dp[i+1][3]=mod(dp[i+1][3],dp[i][3]);}else if(k<l){  // ab ba bcdp[i+1][3]=mod(dp[i+1][3],dp[i][2]);dp[i+1][3]=mod(dp[i+1][3],dp[i][3]);}else if(j==k){ // ac ac bbdp[i+1][1]=mod(dp[i+1][1],dp[i][1]);dp[i+1][3]=mod(dp[i+1][3],dp[i][3]);} else if(k==l){ // ac bb bbdp[i+1][2]=mod(dp[i+1][2],dp[i][2]);dp[i+1][3]=mod(dp[i+1][3],dp[i][3]);} elsedp[i+1][3]=mod(dp[i+1][3],dp[i][3]);} }}}cout<<dp[p][3]<<endl;}return 0;} 


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