AtCoder ABC 130E Common Subsequence

题目链接：https://atcoder.jp/contests/abc130/tasks/abc130_e

题目大意

　　给定一个长度为 N 的序列 S 和一个长度为 M 的序列 T，问 S 和 T 中有多少对不相同的公共子序列？

　　PS：两个子序列对只有同时满足内容完全相同和在 S 和 T 中的位置完全相同才算相同。

分析

　　设 S(i) 表示 S 序列的前 i 项，设 T(j) 表示 T 序列的前 j 项，dp[i][j] 表示 S(i) 和 T(j) 中有多少对不相同的公共子序列。

　　首先 dp[i - 1][j]，dp[i][j - 1], dp[i - 1][j - 1] 有的，dp[i][j] 都有，因此 dp[i][j] 至少是 dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1]。

　　区别在于 S[i] 和 T[j]，如果它们相等，那么 dp[i][j] 就要加上 dp[i - 1][j - 1]，相当于把 dp[i - 1][j - 1] 中的所有对公共子序列都延长了 1；否则不变。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
#define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
#define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

template<class T>
inline string toString(T x) {
    ostringstream sout;
    sout << x;
    return sout.str();
}

inline int toInt(string s) {
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}

//min <= aim <= max
template<typename T>
inline bool BETWEEN(const T aim, const T min, const T max) {
    return min <= aim && aim <= max;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< int, PII > PIPII;
typedef pair< string, int > PSI;
typedef pair< int, PSI > PIPSI;
typedef set< int > SI;
typedef set< PII > SPII;
typedef vector< int > VI;
typedef vector< double > VD;
typedef vector< VI > VVI;
typedef vector< SI > VSI;
typedef vector< PII > VPII;
typedef map< int, int > MII;
typedef map< int, string > MIS;
typedef map< int, PII > MIPII;
typedef map< PII, int > MPIII;
typedef map< string, int > MSI;
typedef map< string, string > MSS;
typedef map< PII, string > MPIIS;
typedef map< PII, PII > MPIIPII;
typedef multimap< int, int > MMII;
typedef multimap< string, int > MMSI;
//typedef unordered_map< int, int > uMII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-8;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 2e3 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

int N, M, S[maxN], T[maxN];
// dp[i][j]表示 S 的前 i 项和 T 的前 j 项有多少相同的子序列对 
LL dp[maxN][maxN];

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    INIT();
    cin >> N >> M;
    For(i, 1, N) cin >> S[i];
    For(i, 1, M) cin >> T[i];
    
    // 初始化（空集算一个） 
    For(i, 0, N) dp[i][0] = 1;
    For(j, 0, M) dp[0][j] = 1;
    
    For(i, 1, N) {
        For(j, 1, M) {
            // 容斥原理 
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
            if(S[i] == T[j]) dp[i][j] += dp[i - 1][j - 1];
            dp[i][j] = (dp[i][j] + mod) % mod;
        }
    }
    
    cout << dp[N][M] << endl;
    return 0;
}