HDU-6044 Limited Permutation（计数）

传送门：点这里~

题意：有n个区间，对于第i个区间[li,ri]有li<=i<=ri,
对于任意1<=L<=i<=R<=n，当前仅当li<=L<=i<=R<=ri时P[i]=min(P[L],P[L+1],...,P[R])

题解：
首先要理解题意：当前仅当li<=L<=i<=R<=ri时P[i]=min(P[L],P[L+1],...,P[R])
因此对于P[i]一定有P[i]>P[li-1]且P[i]>P[ri+1]，进一步说区间[li,ri](除了[1,n])一定被某个区间[lj,rj]包含,且j=li-1或j=ri+1
即区间j可分成[lj,j-1]和[j+1,rj]

我们把n个区间按L升序R降序进行排序(这样得到的区间LR正是前序遍历的区间,区间由大到小)。得到的第1个区间一定要是[1,n](1比任何数都小)，否则不合法，输出0；设这个区间对应的是第i个数，因此区间可再分为[1,i-1]和[i+1,n],看是否有这2个区间，如果没有则不合法，输出0...直到区间不可再分。

现在再来考虑方法数：设f(i)为区间i内的方法数，u,v分别为左右子区间，i内一共有ri-li+1个数，除去中间一个，要从中选i-li个数放入左区间，剩下的放入右区间，因此答案为：f(i)=f(u)*f(v)*C(ri-li,i-li)

#include<bits/stdc++.h>using namespace std;typedef long long LL;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);    }    inline bool read(int &t) {        while(c < sz && buf[c] != '-' && (buf[c] < '0' || buf[c] > '9')) c++;        if(c >= sz) return false;        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;    }}const LL mod = 1e9 + 7;const int MX = 1e6 + 5;LL F[MX], invF[MX];LL power(LL a, LL b) {    LL ret = 1;    while (b) {        if (b & 1) ret = ret * a % mod;        a = a * a % mod;        b >>= 1;    }    return ret;}void init() {    F[0] = 1;    for (int i = 1; i < MX; i++) F[i] = F[i - 1] * i % mod;    invF[MX - 1] = power(F[MX - 1], mod - 2);    for (int i = MX - 2; i >= 0; i--) invF[i] = invF[i + 1] * (i + 1) % mod;}LL C(LL n, LL m) {    LL ret = 1;    while (n && m) {        LL nn = n % mod, mm = m % mod;        if (nn < mm) return 0;        ret = ((ret * F[nn] % mod) * invF[mm] % mod) * invF[nn - mm] % mod;        n /= mod, m /= mod;    }    return ret;}struct node {    int l, r, id;    bool operator<(const node& _A)const {        if (l != _A.l) return l < _A.l;        return r > _A.r;    }} a[MX];int n;int mark;  //是否有区间不合法int rear;LL dfs(int l, int r) {    if (mark == 0) return 0;    if (l > r) return 1;    if (a[rear].l != l || a[rear].r != r) {        mark = 0;        return 0;    }    node now = a[rear++];    LL ret = C(now.r - now.l, now.id - now.l) * dfs(now.l, now.id - 1) % mod;    ret = (ret * dfs(now.id + 1, now.r)) % mod;    return ret;}int main() {    int cas = 0;    init();    //freopen("in.txt", "r", stdin);    IO::begin();    while (IO::read(n)) {        for (int i = 1; i <= n; i++) IO::read(a[i].l);        for (int i = 1; i <= n; i++) IO::read(a[i].r), a[i].id = i;        sort(a + 1, a + n + 1);        mark = rear = 1;        printf("Case #%d: %lld\n", ++cas, dfs(1, n));    }    return 0;}