UVALive 6257 Chemist's vows --一道题的三种解法（模拟，DFS，DP）
题意：给一个元素周期表的元素符号（114种），再给一个串，问这个串能否有这些元素符号组成（全为小写）。
解法1：动态规划
定义：dp[i]表示到 i 这个字符为止，能否有元素周期表里的符号构成。
则有转移方程：dp[i] = (dp[i-1]&&f(i-1,1)) || (dp[i-2]&&f(i-2,2)) f(i,k):表示从i开始填入k个字符，这k个字符在不在元素周期表中。 dp[0] = 1
代码：
//109ms 0KB
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
#define N 50007 string single[] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"};
string ss[] = {"he","li","be","ne","na","mg",
"al","si","cl","ar","ca","sc","ti","cr","mn",
"fe","co","ni","cu","zn","ga","ge","as","se",
"br","kr","rb","sr","zr","nb","mo","tc","ru",
"rh","pd","ag","cd","in","sn","sb","te","xe",
"cs","ba","hf","ta","re","os","ir","pt","au",
"hg","tl","pb","bi","po","at","rn","fr","ra",
"rf","db","sg","bh","hs","mt","ds","rg","cn",
"fl","lv","la","ce","pr","nd","pm","sm","eu",
"gd","tb","dy","ho","er","tm","yb","lu","ac",
"th","pa","np","pu","am","cm","bk","cf","es",
"fm","md","no","lr"}; int vis[][],tag[];
int dp[N];
char st[N]; void init()
{
memset(vis,,sizeof(vis));
memset(tag,,sizeof(tag));
for(int i=;i<;i++)
tag[single[i][]-'a'] = ;
for(int i=;i<;i++)
vis[ss[i][]-'a'][ss[i][]-'a'] = ;
} int main()
{
int t,len,i;
init();
scanf("%d",&t);
while(t--)
{
scanf("%s",st+);
len = strlen(st+);
memset(dp,,sizeof(dp));
dp[] = ;
for(i=;i<len;i++)
{
if(dp[i])
{
if(tag[st[i+]-'a'])
dp[i+] = ;
dp[i+] |= vis[st[i+]-'a'][st[i+]-'a'];
}
}
if(dp[len])
puts("YES");
else
puts("NO");
}
return ;
}
解法2：DFS
搜索时循环的是元素周期表的符号个数。详见代码
代码： （306ms）
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
#define N 50007 string ss[] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v","he","li","be","ne","na","mg",
"al","si","cl","ar","ca","sc","ti","cr","mn",
"fe","co","ni","cu","zn","ga","ge","as","se",
"br","kr","rb","sr","zr","nb","mo","tc","ru",
"rh","pd","ag","cd","in","sn","sb","te","xe",
"cs","ba","hf","ta","re","os","ir","pt","au",
"hg","tl","pb","bi","po","at","rn","fr","ra",
"rf","db","sg","bh","hs","mt","ds","rg","cn",
"fl","lv","la","ce","pr","nd","pm","sm","eu",
"gd","tb","dy","ho","er","tm","yb","lu","ac",
"th","pa","np","pu","am","cm","bk","cf","es",
"fm","md","no","lr"}; int vis[N];
int len[];
char st[N];
int Length;
bool Tag; void init()
{
int i;
for(i=;i<;i++)
len[i] = ;
for(i=;i<;i++)
len[i] = ;
} void dfs(int u)
{
if(u == Length)
Tag = ;
if(Tag)
return;
for(int i=;i<;i++)
{
int flag = ;
if(u+len[i] <= Length && !vis[u+len[i]])
{
for(int j=;j<len[i];j++)
{
if(ss[i][j] != st[u+j])
{
flag = ;
break;
}
}
if(flag)
{
vis[u+len[i]] = ;
dfs(u+len[i]);
}
}
}
} int main()
{
init();
int t,i;
scanf("%d",&t);
while(t--)
{
scanf("%s",st);
Length = strlen(st);
memset(vis,,sizeof(vis));
Tag = ;
dfs();
if(Tag)
puts("YES");
else
puts("NO");
}
return ;
}
解法3：乱搞，模拟。
分成： 单个元素存在与否，与前面匹不匹配，与后面匹不匹配，总共2^3 = 8种情况，然后O(n)扫过去，代码很长。。。
代码：（586ms）
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
#define N 50007 string single[] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"};
string ss[] = {"he","li","be","ne","na","mg",
"al","si","cl","ar","ca","sc","ti","cr","mn",
"fe","co","ni","cu","zn","ga","ge","as","se",
"br","kr","rb","sr","zr","nb","mo","tc","ru",
"rh","pd","ag","cd","in","sn","sb","te","xe",
"cs","ba","hf","ta","re","os","ir","pt","au",
"hg","tl","pb","bi","po","at","rn","fr","ra",
"rf","db","sg","bh","hs","mt","ds","rg","cn",
"fl","lv","la","ce","pr","nd","pm","sm","eu",
"gd","tb","dy","ho","er","tm","yb","lu","ac",
"th","pa","np","pu","am","cm","bk","cf","es",
"fm","md","no","lr"}; char st[N];
int vis[N]; int main()
{
int t,len,i,j,k;
scanf("%d",&t);
while(t--)
{
scanf("%s",st);
len = strlen(st);
int flag = ;
memset(vis,,sizeof(vis));
for(i=;i<len;i++)
{
if(vis[i])
continue;
string S = "";
S += st[i];
for(j=;j<;j++)
{
if(single[j] == S)
break;
}
if(j == ) //not single
{
if(i > && !vis[i-])
{
S = st[i-]+S;
for(j=;j<;j++)
{
if(ss[j] == S)
break;
}
if(j != ) //pre match
{
if(i < len-)
{
string ks = "";
ks += st[i];
ks += st[i+];
for(k=;k<;k++)
{
if(ss[k] == ks)
break;
}
if(k != ) //back match
vis[i] = ;
else //back not match
vis[i] = ;
}
}
else //pre not match
{
if(i < len-)
{
string ks = "";
ks += st[i];
ks += st[i+];
for(k=;k<;k++)
{
if(ss[k] == ks)
break;
}
if(k != ) //back match
vis[i+] = ;
else //back not match
{
flag = ;
break;
}
}
else
{
flag = ;
break;
}
}
}
else
{
if(i < len-)
{
string ks = "";
ks += st[i];
ks += st[i+];
for(k=;k<;k++)
{
if(ss[k] == ks)
break;
}
if(k != ) //back match
vis[i+] = ;
else //back not match
{
flag = ;
break;
}
}
else
{
flag = ;
break;
}
}
}
else //single
{
if(i > && !vis[i-])
{
S = st[i-]+S;
for(j=;j<;j++)
{
if(ss[j] == S)
break;
}
if(j != ) //pre match
{
if(i < len-)
{
string ks = "";
ks += st[i];
ks += st[i+];
for(k=;k<;k++)
{
if(ss[k] == ks)
break;
}
if(k != ) //back match
vis[i] = ;
else //back not match
vis[i] = ;
}
}
else //pre not match
{
if(i < len-)
{
string ks = "";
ks += st[i];
ks += st[i+];
for(k=;k<;k++)
{
if(ss[k] == ks)
break;
}
if(k != ) //back match
vis[i] = ;
else //back not match
vis[i] = ;
}
else
vis[i] = ;
}
}
else
{
if(i < len-)
{
string ks = "";
ks += st[i];
ks += st[i+];
for(k=;k<;k++)
{
if(ss[k] == ks)
break;
}
if(k != ) //back match
vis[i] = ;
else //back not match
vis[i] = ;
}
else
vis[i] = ;
}
}
}
if(flag)
puts("YES");
else
puts("NO");
}
return ;
}
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