剪枝:
1. 对于任意串, 任意相邻的'C', 'O', 'W'之间的串应该在目标串中能找到.
2. 通过ELFHash对串进行判重.( HASHSIZE可以大点)
3. 若串的长度length 减去 目标串长度(47) MOD 3不等于0, 排除.
/**//*
ID: lorelei3
TASK: cryptcow
LANG: C++
*/
#include <iostream>
#include <memory.h>
#include <fstream>
#include <string>
using namespace std;
const int MAX = 100;
const int MOD = 131071;
const int LEN = 47;
const string plaintext = "Begin the Escape execution at the Break of Dawn";
ifstream fin("cryptcow.in");
ofstream fout("cryptcow.out");
bool visited[MOD+1];
inline bool ELFHash(string s, int length)
{
unsigned long h=0,g;
for (int i=0; i<length; i++){
h=(h<<4)+s[i];
g=h&0xf0000000l;
if(g){
h^=g>>24;
}
h&=~g;
}
h = h%MOD;
if(visited[h])
return true;
else{
visited[h] = true;
return false;
}
}
void dfs(string s, int num){
if(plaintext == s){
fout<<1<<" "<<num<<endl;
exit(0);
}
int length = s.length();
if(ELFHash(s, length)) return ;
int cnt=0, c_cnt=0, o_cnt=0, w_cnt=0;
int c[MAX], o[MAX], w[MAX], label[MAX];
memset(c, 0, sizeof(c));
memset(o, 0, sizeof(o));
memset(w, 0, sizeof(w));
memset(label, 0, sizeof(label));
int i,j,k;
label[cnt++] = -1;
for(i = 0; i<length; ++i){
if(s[i] == 'C'){
c[c_cnt++] = i;
label[cnt++] = i;
}else if(s[i] == 'O'){
o[o_cnt++] = i;
label[cnt++] = i;
}else if(s[i] == 'W'){
w[w_cnt++] = i;
label[cnt++] = i;
}
}
label[cnt++] = length;
if(c_cnt!=o_cnt || o_cnt!=w_cnt)
return;
for(i=0; i<cnt-1; ++i){
if(label[i]+1 < label[i+1]){
string sub = s.substr(label[i]+1, label[i+1]-label[i]-1);
if(plaintext.find(sub)==string::npos)
return;
}
}
for(i=0; i<o_cnt; ++i){
for(j=0; j<c_cnt; ++j){
for(k=w_cnt-1; k>=0; --k){
if(c[j]<o[i] && o[i]<w[k]){
string t1 = s.substr(0, c[j]);
string t2 = s.substr(o[i]+1, w[k]-o[i]-1);
string t3 = s.substr(c[j]+1, o[i]-c[j]-1);
string t4 = s.substr(w[k]+1, length-w[k]);
string t = t1+t2+t3+t4;
dfs(t, num+1);
}
}
}
}
}
int main(){
string s;
getline(fin, s);
int length = s.length();
if( (length-LEN)%3 !=0 ){
fout<<"0 0"<<endl;
return 0;
}
dfs(s,0);
fout<<"0 0"<<endl;
return 0;
}
posted on 2011-01-24 00:25
小阮 阅读(692)
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