研究一下建树 ：

/*HDU 6044 - Limited Permutation [ 读入优化,笛卡尔树 ]  |  2017 Multi-University Training Contest 1题意：	给出两组序列 l[i], r[i]， 代表以 p[i] 为最小值所在的区间的边界	问 满足这些条件的序列 p 的个数分析：	必定能找到一个p[i] 使得 l[i] == 1, r[i] == n ，其将数组分成两块[1, i-1], [i+1, n]	以之为根节点，将区间为[1, i-1] 和 [i+1, n] 的节点作为左右儿子，再在每一块上递归地进行划分，则构成一棵笛卡尔树	能构成的笛卡尔树的形状是唯一的，若不能构成，则 ans = 0	若能构成，则 以i为根的子树的方案数 f(i) = C(size_l+size_r, size_l) * f(l) * f(r)	数据很多用fread读入*/#include 
     
      using namespace std;#define LL long longconst int N = 1e6+5;const LL MOD = 1e9+7;namespace IO {    const int MX = 4e7; //1e7 占用内存 11000kb    char buf[MX]; int c, sz;    void begin() {        c = 0;        sz = fread(buf, 1, MX, stdin);//一次性全部读入    }    inline bool read(int &t) {        while (c < sz && buf[c] != '-' && (buf[c] < '0' || buf[c] > '9')) c++;        if (c >= sz) return false;//若读完整个缓冲块则退出        bool flag = 0; if(buf[c] == '-') flag = 1, c++;        for(t = 0; c < sz && '0' <= buf[c] && buf[c] <= '9'; c++) t = t * 10 + buf[c] - '0';        if(flag) t = -t;        return true;    }}namespace COMB {	int F[N], Finv[N], inv[N];//F是阶乘，Finv是逆元的阶乘	void init(){		inv[1] = 1;		for(int i = 2; i < N; i ++){			inv[i] = (MOD - MOD / i) * 1LL * inv[MOD % i] % MOD;		}		F[0] = Finv[0] = 1;		for(int i = 1; i < N; i ++){			F[i] = F[i-1] * 1LL * i % MOD;			Finv[i] = Finv[i-1] * 1LL * inv[i] % MOD;		}	}	int comb(int n, int m){//comb(n, m)就是C(n, m)		if(m < 0 || m > n) return 0;		return F[n] * 1LL * Finv[n - m] % MOD * Finv[m] % MOD;	}}struct Node{    int f, l, r;}T[N];int l[N], r[N];int n;int q[N], top;inline bool cmp(const int& a,const int& b) {    return (r[a] - l[a]) < (r[b] - l[b]);}int Build(){    top = 0;    for (int i = 1; i <= n; i++)    {        int k = top;        while (k && cmp(q[k-1], i)) --k;        if (k != 0)        {            T[i].f = q[k-1];            T[q[k-1]].r = i;        }        if (k != top)        {            T[q[k]].f = i;            T[i].l = q[k];        }        q[k++] = i;        top = k;    }    return q[0];}bool flag;LL ans;LL dfs(int x, int L, int R){    if (!x) return 1;    if (!flag) return 0;    if (l[x] != L || r[x] != R) return 0;    LL res = COMB::comb(R-L, R-x);    res = res * dfs(T[x].l, L, x-1) % MOD;    res = res * dfs(T[x].r, x+1, R) % MOD;    if (res == 0) flag = 0;    return res;}int main(){    COMB::init();    IO::begin();    for (int tt = 1; IO::read(n); ++tt)    {        for (int i = 1; i <= n; i++) T[i].l = T[i].r = T[i].f = 0;        for (int i = 1; i <= n; i++) IO::read(l[i]);        for (int i = 1; i <= n; i++) IO::read(r[i]);        int root = Build();        flag = 1;        ans = dfs(root, 1, n);        printf("Case #%d: %lld\n", tt, ans);    }}
     

　　要么直接 map

#include 
     
      using namespace std;#define LL long longconst int N = 1e6+5;const LL MOD = 1e9+7;namespace fastIO {	#define BUF_SIZE 100000	//fread -> read	bool IOerror = 0;	inline char nc() {		static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;		if(p1 == pend) {			p1 = buf;			pend = buf + fread(buf, 1, BUF_SIZE, stdin);			if(pend == p1) {				IOerror = 1;				return -1;			}		}		return *p1++;	}	inline bool blank(char ch) {		return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';	}	inline void read(int &x) {		char ch;		while(blank(ch = nc()));		if(IOerror)			return;		for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');	}	#undef BUF_SIZE}namespace COMB {	int F[N], Finv[N], inv[N];//F是阶乘，Finv是逆元的阶乘	void init(){		inv[1] = 1;		for(int i = 2; i < N; i ++){			inv[i] = (MOD - MOD / i) * 1LL * inv[MOD % i] % MOD;		}		F[0] = Finv[0] = 1;		for(int i = 1; i < N; i ++){			F[i] = F[i-1] * 1LL * i % MOD;			Finv[i] = Finv[i-1] * 1LL * inv[i] % MOD;		}	}	int comb(int n, int m){//comb(n, m)就是C(n, m)		if(m < 0 || m > n) return 0;		return F[n] * 1LL * Finv[n - m] % MOD * Finv[m] % MOD;	}}typedef pair
      
        P;using namespace fastIO;map
       
         mp;int t, n;int l[N], r[N];LL dfs(int l, int r){    if (l > r) return 1;    int x = mp[P(l, r)];    if (!x) return 0;    LL ans = COMB::comb(r-l, r-x);    ans = ans * dfs(l, x-1) % MOD;    ans = ans * dfs(x+1, r) % MOD;    return ans;}int main(){    COMB::init();    for (int tt = 1; read(n), !IOerror; ++tt)    {        mp.clear();        for (int i = 1; i <= n; i++) read(l[i]);        for (int i = 1; i <= n; i++) read(r[i]);        for (int i = 1; i <= n; i++)            mp[P(l[i], r[i])] = i;        LL ans = dfs(1, n);        printf("Case #%d: %lld\n", tt, ans);    }}