HDU

               思路：对于每个字符，如果它能被替换一定要优先替换，其次再进行删除。遵循这个策略即可。

证明：

对于这题的第一个测试数据：

abbaaddba 1d b

当匹配到'b'  和 'd'时应该优先替换而不是删除'd'，这样可以保证不删除掉有用的字符。

AC代码

#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;typedef vector<char> v;const int maxn = 1000 + 5;map<char, v>ha;char a[maxn], b[maxn];int main() {int T, n, m, k, kase = 1;scanf("%d", &T);while(T--) {ha.clear();scanf("%s%s", a, b);scanf("%d", &k);getchar();char x, y;for(int i = 0; i < k; ++i) {scanf("%c %c", &x, &y);getchar();ha[x].push_back(y);}printf("Case #%d: ", kase++);n = strlen(a), m = strlen(b);int flag = 0, i, j;for(i = 0, j = 0; i < n && j < m; ++j) {if(a[i] != b[j]) {int ok = 0;v &u = ha[b[j]];for(int h = 0; h < u.size(); ++h) {if(u[h] == a[i]) {ok = 1;break;}}if(ok) ++i;}else if(a[i] == b[j]) ++i;}if(j <= m && i == n) flag = 1;if(flag) printf("happy\n");else printf("unhappy\n");}return 0;} 

如有不当之处欢迎指出！