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; }