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[25] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"};
string ss[130] = {"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[30][30],tag[30];
int dp[N];
char st[N];
void init()
{
memset(vis,0,sizeof(vis));
memset(tag,0,sizeof(tag));
for(int i=0;i<14;i++)
tag[single[i][0]-‘a‘] = 1;
for(int i=0;i<100;i++)
vis[ss[i][0]-‘a‘][ss[i][1]-‘a‘] = 1;
}
int main()
{
int t,len,i;
init();
scanf("%d",&t);
while(t--)
{
scanf("%s",st+1);
len = strlen(st+1);
memset(dp,0,sizeof(dp));
dp[0] = 1;
for(i=0;i<len;i++)
{
if(dp[i])
{
if(tag[st[i+1]-‘a‘])
dp[i+1] = 1;
dp[i+2] |= vis[st[i+1]-‘a‘][st[i+2]-‘a‘];
}
}
if(dp[len])
puts("YES");
else
puts("NO");
}
return 0;
}
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[130] = {"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[140];
char st[N];
int Length;
bool Tag;
void init()
{
int i;
for(i=0;i<14;i++)
len[i] = 1;
for(i=14;i<114;i++)
len[i] = 2;
}
void dfs(int u)
{
if(u == Length)
Tag = 1;
if(Tag)
return;
for(int i=0;i<114;i++)
{
int flag = 1;
if(u+len[i] <= Length && !vis[u+len[i]])
{
for(int j=0;j<len[i];j++)
{
if(ss[i][j] != st[u+j])
{
flag = 0;
break;
}
}
if(flag)
{
vis[u+len[i]] = 1;
dfs(u+len[i]);
}
}
}
}
int main()
{
init();
int t,i;
scanf("%d",&t);
while(t--)
{
scanf("%s",st);
Length = strlen(st);
memset(vis,0,sizeof(vis));
Tag = 0;
dfs(0);
if(Tag)
puts("YES");
else
puts("NO");
}
return 0;
}
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[25] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"};
string ss[130] = {"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 = 1;
memset(vis,0,sizeof(vis));
for(i=0;i<len;i++)
{
if(vis[i])
continue;
string S = "";
S += st[i];
for(j=0;j<14;j++)
{
if(single[j] == S)
break;
}
if(j == 14) //not single
{
if(i > 0 && !vis[i-1])
{
S = st[i-1]+S;
for(j=0;j<100;j++)
{
if(ss[j] == S)
break;
}
if(j != 100) //pre match
{
if(i < len-1)
{
string ks = "";
ks += st[i];
ks += st[i+1];
for(k=0;k<100;k++)
{
if(ss[k] == ks)
break;
}
if(k != 100) //back match
{
vis[i] = 0;
}
else //back not match
vis[i] = 1;
}
}
else //pre not match
{
if(i < len-1)
{
string ks = "";
ks += st[i];
ks += st[i+1];
for(k=0;k<100;k++)
{
if(ss[k] == ks)
break;
}
if(k != 100) //back match
{
vis[i+1] = 1;
}
else //back not match
{
flag = 0;
break;
}
}
else
{
flag = 0;
break;
}
}
}
else
{
if(i < len-1)
{
string ks = "";
ks += st[i];
ks += st[i+1];
for(k=0;k<100;k++)
{
if(ss[k] == ks)
break;
}
if(k != 100) //back match
{
vis[i+1] = 1;
}
else //back not match
{
flag = 0;
break;
}
}
else
{
flag = 0;
break;
}
}
}
else //single
{
if(i > 0 && !vis[i-1])
{
S = st[i-1]+S;
for(j=0;j<100;j++)
{
if(ss[j] == S)
break;
}
if(j != 100) //pre match
{
if(i < len-1)
{
string ks = "";
ks += st[i];
ks += st[i+1];
for(k=0;k<100;k++)
{
if(ss[k] == ks)
break;
}
if(k != 100) //back match
{
vis[i] = 0;
}
else //back not match
vis[i] = 1;
}
}
else //pre not match
{
if(i < len-1)
{
string ks = "";
ks += st[i];
ks += st[i+1];
for(k=0;k<100;k++)
{
if(ss[k] == ks)
break;
}
if(k != 100) //back match
{
vis[i] = 0;
}
else //back not match
{
vis[i] = 1;
}
}
else
{
vis[i] = 1;
}
}
}
else
{
if(i < len-1)
{
string ks = "";
ks += st[i];
ks += st[i+1];
for(k=0;k<100;k++)
{
if(ss[k] == ks)
break;
}
if(k != 100) //back match
{
vis[i] = 0;
}
else //back not match
{
vis[i] = 1;
}
}
else
{
vis[i] = 1;
}
}
}
}
if(flag)
puts("YES");
else
puts("NO");
}
return 0;
}
UVALive 6257 Chemist's vows --一道题的三种解法（模拟，DFS，DP）,布布扣,bubuko.com
UVALive 6257 Chemist's vows --一道题的三种解法（模拟，DFS，DP）
时间： 2024-10-12 07:40:58