Light OJ 1013 - Love Calculator(LCS+ 計方案數)

大致題意：有a，b字符串，求最短的字符串使a，b均爲它的子序列，求這種最短字符串有多少個

思路： 顯然最短長度就是｜a｜＋｜b｜－ LCS

同樣dp兩遍，第一遍求LCS，第二遍在LCS的轉以上dp出方案數:

如果a[i] == b[i] , cnt[i][j] += cnt[i-1][j-1];

否則，有兩種策略，a[i]是末尾最後一個字符，或b[j]是最後一個字符

複雜度O（n^2）

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
#define rep(i,n) for ( int i=0; i< int(n); i++ )
using namespace std;
typedef long long ll;
#define X first
#define Y second
#define PB push_back
#define MP make_pair
typedef pair<int,int> pii;

template <class T>
inline bool RD(T &ret) {
        char c; int sgn;
        if (c = getchar(), c == EOF) return 0;
        while (c != '-' && (c<'0' || c>'9')) c = getchar();
        sgn = (c == '-') ? -1 : 1 , ret = (c == '-') ? 0 : (c - '0');
        while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
        ret *= sgn;
        return 1;
}
template <class T>
inline void PT(T x) {
        if (x < 0) putchar('-') ,x = -x;
        if (x > 9) PT(x / 10);
        putchar(x % 10 + '0');
}
const int N = 33;
int minn[N][N];
ll cnt[N][N];
char a[N], b[N];
int lena, lenb;

int main() {
	int T;
	scanf("%d", &T);
	int cas = 0;
	while( T -- ) {	
		scanf("%s%s", a + 1, b + 1);
		lena = strlen(a + 1);
		lenb = strlen(b + 1);
		memset( minn , 0, sizeof( minn) );
		memset( cnt, 0, sizeof(cnt));
		REP(i, lena) minn[i][0] = i;
		REP(i, lenb) minn[0][i] = i;
		REP(i, lena) {
			REP(j, lenb) {
				minn[i][j] = min( min( minn[i][j-1] + 1, minn[i-1][j] + 1 ), minn[i-1][j-1] + 2 - (a[i] == b[j]) );
			}
		}
		rep(i, lenb + 1) cnt[0][i] = 1;
		rep(i, lena + 1) cnt[i][0] = 1;
		REP(i, lena) {
			REP(j, lenb) {
				if( a[i] == b[j] ) cnt[i][j] += cnt[i-1][j-1];
				else {
					if( minn[i][j] == minn[i-1][j] + 1 ) cnt[i][j] += cnt[i-1][j];
					if( minn[i][j] == minn[i][j-1] + 1 ) cnt[i][j] += cnt[i][j-1];
				}
			}
		}
		printf("Case %d: %d %lld\n", ++ cas, minn[lena][lenb], cnt[lena][lenb] );
	}
	
}