题面
题解
字符很麻烦,不妨用数字代替(比如1代表'W')
const char c[5] = {0, 'W', 'I', 'N', 'G'};
接着,像这种两个子串可以合并成另一个子串的题可以考虑区间$DP$
设$bool$数组$f_{i,j,k}$表示区间$[l,r]$能否合成单个字符$c_k$
于是就可以套区间$DP$的板子了
#include <map>
#include <cstdio>
#include <cstring>
#include <algorithm>
using std::min; using std::max;
using std::sort; using std::swap;
using std::unique; using std::lower_bound;
using std::map;
typedef long long ll;
template<typename T>
void read(T &x) {
int flag = 1; x = 0; char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}
const int Len = 2e2 + 10, _ = 5;
const char c[5] = {0, 'W', 'I', 'N', 'G'};
char s[Len]; bool p, f[Len][Len][_];
int num[_], list[Len][_], tot, len;
int check(char x) { for(int i = 1; i <= 4; ++i) if(x == c[i]) return i; }
int main () {
for(int i = 1; i <= 4; ++i) read(num[i]);
for(int i = 1; i <= 4; ++i)
for(int j = 1; j <= num[i]; ++j) {
char ss[5]; scanf("%s", ss);
list[++tot][0] = i;
list[tot][1] = check(ss[0]);
list[tot][2] = check(ss[1]);
}
scanf("%s", s + 1), len = strlen(s + 1);
for(int i = 1; i <= len; ++i) f[i][i][check(s[i])] = true;
for(int k = 1; k <= len; ++k)
for(int i = 1; i + k <= len; ++i) {
int j = i + k;
for(int p = i; p < j; ++p)
for(int l = 1; l <= tot; ++l)
if(f[i][p][list[l][1]] && f[p + 1][j][list[l][2]])
f[i][j][list[l][0]] = true;
}
for(int i = 1; i <= 4; ++i)
if(f[1][len][i]) putchar(c[i]), p = true;
if(!p) printf("The name is wrong!");
puts("");
return 0;
}