USACO 4.1 Cryptcowgraphy(DFS)

这个题 我看了题解的提示，把自己的代码弄的只过了8组。。。最后两组无解的情况，实在搜不出来了。

加了判字符串哈希的数组，就是为了搞最后两组数组的。。。终于水过了。。。

代码非常难看，Tire树,字符串哈希（找的别人的）。。。

View Code

/*
ID: cuizhe
LANG: C++
TASK: cryptcow
*/
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <ctime>
using namespace std;
#define MOD 99997
char str[101],z;
int oo[240];
char aim[100] = "Begin the Escape execution at the Break of Dawn";
int tire[10000][30];
int t = 0;
int time1,time2;
int hash[MOD];
int elfhash(char * str) {

unsigned int res = 0,g;
while (*str)
{
res = (res << 4) + (*str++);
g = res & 0xF0000000;
if (g) res ^= g >> 24;
res &= ~g;
}
return res % MOD;
}
int find(char *s)
{
int root = 0,len,i;
len = strlen(s);
if(len == 0) return 1;
for(i = 0; i < len; i ++)
{
if(!tire[root][oo[s[i]]])
{
return 0;
}
root = tire[root][oo[s[i]]];
}
return 1;
}
void build(char *s)
{
int i,len,root;
len = strlen(s);
root = 0;
for(i = 0; i < len; i ++)
{
if(!tire[root][oo[s[i]]])
{
tire[root][oo[s[i]]] = t ++;
root = t;
}
else
{
root = tire[root][oo[s[i]]];
}
}
}
void dfs(int step)
{
int i,len,j,k,u,v,flag;
char s[101];
char temp[101];
int qu1[10],qu2[10],qu3[10];
int f1,f2,f3;
f1 = f2 = f3 = 0;
if(strcmp(str,aim) == 0)
{
z = 1;
return ;
}
if(step == 0)
return ;
int kk;
if(hash[kk = elfhash(str)] < 3)
hash[kk] ++;
else
return ;
strcpy(temp,str);
if(z) return ;
len = strlen(str);
flag = 0;
for(i = 0; i < len; i ++)
{
if(str[i] == 'C')
{
s[flag] = '\0';
if(!find(s))
{
return;
}
qu1[f1++] = i;
flag = 0;
}
else if(str[i] == 'O')
{
s[flag] = '\0';
if(!find(s))
{
return;
}
if(f1 == 0) return;
qu2[f2++] = i;
flag = 0;
}
else if(str[i] == 'W')
{
s[flag] = '\0';
if(!find(s))
{
return;
}
if(f1 == 0) return;
qu3[f3++] = i;
flag = 0;
}
else
{
s[flag ++] = str[i];
}
}
for(i = 0; i < qu1[0]; i ++)
{
if(aim[i] != str[i])
return ;
}
for(j = 0; j < f2; j ++)
{
for(i = 0; i < f1; i ++)
{
for(k = f3-1; k >= 0; k --)
{
if(qu1[i] < qu2[j] &&qu2[j] < qu3[k])
{
for(u = qu1[i],v = qu2[j]+1; v < qu3[k]; u ++,v ++)
{
str[u] = temp[v];
}
for(v = qu1[i]+1; v < qu2[j]; v ++,u ++)
{
str[u] = temp[v];
}
for(v = qu3[k]+1; v < len; v ++,u ++)
{
str[u] = temp[v];
}
str[u] = '\0';
dfs(step-1);
strcpy(str,temp);
}
}
}
}
}
int main()
{
char s[101];
char ch[101];

int len,i,j,k,num = 0;
int c,o,w;
c = o = w = 0;
z = 0;
freopen("cryptcow.in","r",stdin);
freopen("cryptcow.out","w",stdout);
len = strlen(aim);
for(i = 0; i < len; i ++)
{
if(!oo[aim[i]])
oo[aim[i]] = num ++;
}
for(i = 0; i < len; i ++)
{
for(j = i; j < len; j ++)
{
for(k = i; k <= j; k ++)
{
ch[k-i] = aim[k];
}
ch[k-i] = '\0';
build(ch);
}
}
i = 0;
while(scanf("%c",&s[i])!=EOF)
{
if(s[i] == '\n')
{
len = i;
break;
}
i ++;
}
s[i] = '\0';
len = i;
for(i = 0; i < len; i ++)
{
if(s[i] == 'C')
c ++;
else if(s[i] == 'O')
o ++;
else if(s[i] == 'W')
w ++;
}
strcpy(str,s);
if(c == o&&o == w)
{
dfs(c);
if(z)
printf("1 %d\n",c);
else
printf("0 0\n");
}
else
printf("0 0\n");
return 0;
}