UVA

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思路：dp(i, j)表示第一个串前i个字符和第二个串前j个字符需要的最短字符串长度，cnt(i, j)表示第一个串前i个字符和第二个串前j个字符需要的最短字符串的个数。

转移方程：

if(s1[i] == s2[j]) dp[i][j] = dp[i-1][j-1] + 1;else dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1]);

if(s1[i] == s2[j]) cnt[i][j] = cnt[i-1][j-1]; //字符成功匹配else {if(dp[i-1][j] == dp[i][j-1])         cnt[i][j] = cnt[i-1][j] + cnt[i][j-1];else if(dp[i-1][j] < dp[i][j-1]) cnt[i][j] = cnt[i-1][j];else cnt[i][j] = cnt[i][j-1];}

AC代码

#include <cstdio>#include <cmath>#include <cctype>#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;const int maxn = 30 + 5;char s1[maxn], s2[maxn];int dp[maxn][maxn], cnt[maxn][maxn];int main() {int T, kase = 1;scanf("%d", &T);getchar();while(T--) {fgets(s1+1, sizeof(s1), stdin);fgets(s2+1, sizeof(s2), stdin);//scanf("%s%s", s1+1, s2+1);int n = strlen(s1+1) - 1, m = strlen(s2+1) - 1;for(int i = 0; i < maxn; ++i) {dp[0][i] = dp[i][0] = i;cnt[0][i] = cnt[i][0] = 1;}for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) {if(s1[i] == s2[j]) dp[i][j] = dp[i-1][j-1] + 1;else dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1]);}for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) {if(s1[i] == s2[j]) cnt[i][j] = cnt[i-1][j-1];else {if(dp[i-1][j] == dp[i][j-1]) cnt[i][j] = cnt[i-1][j] + cnt[i][j-1];else if(dp[i-1][j] < dp[i][j-1]) cnt[i][j] = cnt[i-1][j];else cnt[i][j] = cnt[i][j-1];}}printf("Case #%d: %d %d\n", kase++, dp[n][m], cnt[n][m]);}return 0;}

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