UVa 11775 Unique Story

题意：

题意就是给你两个序列，问你最多可以有多少个互不相同的子序列。可以反过来想，求相同的有多少个，所有的子序列一共有2的n次方个（排列组合公式求和）。一看就是DP，转移方程：dp[i][j]=(sum[i-1][j-1]+dp[i][j]+1)%MOD;dp[i][j]表示以i和j结尾的相同的字符串有多少个，sum[i][j]表示以前i个和前j个能够组成的相同的子序列有多少个。对于这个方程的转移，朴素的不行会T，所以我们要用到二维树状数组。具体看代码。

代码：

////  Created by  CQU_CST_WuErli//  Copyright (c) 2015 CQU_CST_WuErli. All rights reserved.//// #include<bits/stdc++.h>#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <list>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <sstream>#define CLR(x) memset(x,0,sizeof(x))#define OFF(x) memset(x,-1,sizeof(x))#define MEM(x,a) memset((x),(a),sizeof(x))#define ALL(x) x.begin(),x.end()#define AT(i,v) for (auto &i:v)#define For_UVa if (kase!=1) cout << endl#define BUG cout << "I am here" << endl#define lookln(x) cout << #x << "=" << x << endl#define look(x) cout << #x << "=" << x#define SI(a) scanf("%d",&a)#define SII(a,b) scanf("%d%d",&a,&b)#define SIII(a,b,c) scanf("%d%d%d",&a,&b,&c)#define Lson l,mid,rt<<1#define Rson mid+1,r,rt<<1|1#define Root 1,n,1#define BigInteger bigntemplate <typename T> T max(T& a,T& b) {return a>b?a:b;}template <typename T> T min(T& a,T& b) {return a<b?a:b;}int gcd(int a,int b) {return b==0?a:gcd(b,a%b);}long long gcd (long long a,long long b) {return b==0LL?a:gcd(b,a%b);}const int MAX_L=2005;// For BigIntegerconst int INF_INT=0x3f3f3f3f;const long long INF_LL=0x7fffffff;const int MOD=10000007;const double eps=1e-9;const double pi=acos(-1);typedef long long  ll;using namespace std;const int N=2000;map<string,int> mp;int ID;vector<int> ss,tt;void process(string &s,vector<int> &vc) {    vc.clear();    vc.push_back(INF_INT);    for (int i=0;i<s.size();i++) {        if (s[i]>='A' && s[i]<='Z') {            string tmp;            tmp+=s[i];            while (s[i+1]>='0' && s[i+1]<='9') tmp+=s[i+1],i++;            if (!mp.count(tmp)) {//              cout << tmp << endl;                mp[tmp]=ID++;            }            vc.push_back(mp[tmp]);        }    }}ll dp[N][N];ll sum[N][N];ll tree[N][N]; int lowbit(int x) {    return x&(-x);}void update(int x,int y,ll val) {    for (int i=x;i<ss.size();i+=lowbit(i))         for (int j=y;j<tt.size();j+=lowbit(j))            tree[i][j]=(tree[i][j]+val)%MOD;}ll query(int x,int y) {    ll ret=0;    for (int i=x;i>0;i-=lowbit(i))         for (int j=y;j>0;j-=lowbit(j))            ret=(ret+tree[i][j])%MOD;    return ret;}ll ksm(int a,int n) {    if (n==0) return 1;    ll ans=ksm(a,n/2);    ans=(ans*ans)%MOD;    if (n%2) ans=(ans*a)%MOD;    return ans;}int main(){#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);//  freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    int T_T;    for (int kase=scanf("%d",&T_T);kase<=T_T;kase++) {        string s,t;        cin >> s >> t;        mp.clear();        ID=1;        process(s,ss);process(t,tt);//      for (int i=1;i<ss.size();i++) cout << ss[i]<< ' ';//      cout << endl;//      for (int i=1;i<tt.size();i++) cout << tt[i] << ' ' ;//      cout << endl;        CLR(dp);CLR(sum);CLR(tree);        for (int i=1;i<ss.size();i++) {            for (int j=1;j<tt.size();j++) {                if (ss[i]==tt[j]){                    dp[i][j]=(sum[i-1][j-1]+dp[i][j]+1)%MOD;                    update(i,j,dp[i][j]);                }                sum[i][j]=query(i,j);            }        }        int Sum=0;//      cout << ksm(2,10) << endl;        Sum=(Sum+ksm(2,ss.size()-1))%MOD;        Sum=(Sum-1+MOD)%MOD;        Sum=(Sum+ksm(2,tt.size()-1))%MOD;        Sum=(Sum-1+MOD)%MOD;//      lookln(Sum);        int ans=(Sum-sum[ss.size()-1][tt.size()-1]*2+2*MOD)%MOD;        cout << "Case " << kase << ": " << ans << endl;     }     return 0;}