Light OJ 1013 - Love Calculator(LCS+ 计方案数)

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大致题意：有a，b字符串，求最短的字符串使a，b均为它的子序列，求这种最短字符串有多少个

思路：显然最短长度就是｜a｜＋｜b｜－ LCS

同样dp两遍，第一遍求LCS，第二遍在LCS的转以上dp出方案数:

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

否则，有两种策略，a[i]是末尾最后一个字符，或b[j]是最后一个字符

复杂度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_pairtypedef 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] );}}

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