luoguP3255 [JLOI2013]地形生成 动态规划

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
    #define re register
    #define de double
    #define le long double
    #define ri register int
    #define ll long long
    #define sh short
    #define pii pair<int, int>
    #define mp make_pair
    #define pb push_back
    #define tpr template <typename ra>
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)    
    extern inline char gc() {
        static char RR[23456], *S = RR + 23333, *T = RR + 23333;
        if(S == T) fread(RR, 1, 23333, stdin), S = RR;
        return *S ++;
    }
    inline int read() {
        int p = 0, w = 1; char c = gc();
        while(c > ‘9‘ || c < ‘0‘) { if(c == ‘-‘) w = -1; c = gc(); }
        while(c >= ‘0‘ && c <= ‘9‘) p = p * 10 + c - ‘0‘, c = gc();
        
        return p * w;
    }
    int wr[50], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = ‘\n‘) {
        if(!o) pc(‘0‘);
        if(o < 0) o = -o, pc(‘-‘);
        while(o) wr[++ rw] = o % 10, o /= 10;
        while(rw) pc(wr[rw --] + ‘0‘);
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; } 
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b, 1 : 0; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b, 1 : 0; }
}
using namespace std;
using namespace remoon;

#define sid 1050
#define mod 2011

inline void inc(int &a, int b) { a += b; if(a >= mod) a -= mod; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }

int n;
struct mountain {
    int h, k;
    friend bool operator < (mountain a, mountain b)
    { return a.h > b.h && (a.h == b.h && a.k < b.k); }
} mt[sid];


int f[1050][1050];
inline void Solve() {
    int ans = 1;
    for(ri i = 1, j = 1; i <= n; i = j + 1) {
        j = i; while(mt[j].h == mt[j + 1].h) j ++;
        rep(ip, i, j) cmin(mt[ip].k, i);
        rep(ip, 1, mt[i].k) f[i][ip] = 1;
        rep(ip, i, j) {
            if(ip < j) rep(jp, 1, mt[ip + 1].k) inc(f[ip][jp], f[ip][jp - 1]);
            else rep(jp, 1, mt[ip].k) inc(f[ip][jp], f[ip][jp - 1]);
            if(ip < j) rep(jp, 1, mt[ip + 1].k) f[ip + 1][jp] = f[ip][jp];
        }
        ans = mul(ans, f[j][mt[j].k]);
    }
    write(ans);
}

int main() {
    
    n = read();
    rep(i, 1, n) mt[i].h = read(), mt[i].k = read();
    sort(mt + 1, mt + n + 1);
    
    int ans = 1, num, pre;
    rep(i, 1, n) {
        if(mt[i].h != mt[i - 1].h) pre = i, num = 0;
        else num ++;
        ans = mul(ans, min(pre, mt[i].k) + num);
    }
    write(ans, ‘ ‘);
    
    Solve();
    return 0;
}