【HDOJ】4343 Interval query

最大不相交集合的数量。
思路是dp[i][j]表示已经有i个不相交集合下一个不相交集合的最右边界。
离散化后，通过贪心解。

/* 4343 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

typedef struct node_t {
    int l, r;
    
    friend bool operator< (const node_t& a, const node_t& b) {
        if (a.l == b.l)
            return a.r < b.r;
        return a.l < b.l;
    }
} node_t;

const int maxn = 1e5+5;
int n, q;
int dp[18][maxn*2];
int X[maxn*2], nx;
node_t nd[maxn];

int getId(int x, bool flag) {
    if (flag)
        return lower_bound(X, X+nx, x) - X;
    else
        return upper_bound(X, X+nx, x) - X - 1;
}

int calc(int l, int r) {
    if (l > r)
        return 0;
    
    if (l==nx || r==-1)
        return 0;
    
    int ret = 0;
    per(i, 0, 18) {
        if (dp[i][l] <= r) {
            ret += (1<<i);
            l = dp[i][l];
        }
    }
    
    return ret;
}

void solve() {
    nx = 0;
    rep(i, 0, n) {
        X[nx++] = nd[i].l;
        X[nx++] = nd[i].r;
    }
    sort(nd, nd+n);
    sort(X, X+nx);
    nx = unique(X, X+nx) - X;
    rep(i, 0, n) {
        nd[i].l = getId(nd[i].l, true);
        nd[i].r = getId(nd[i].r, false);
    }
    
    int mn = nx, l, r = n-1;
    per(i, 0, nx) {
        while (r>=0 && nd[r].l>=i) {
            mn = min(mn, nd[r].r);
            --r;
        }
        dp[0][i] = mn;
    }
    rep(i, 0, 18)
        dp[i][nx] = nx;
    rep(i, 1, 18) {
        rep(j, 0, nx) {
            dp[i][j] = dp[i-1][dp[i-1][j]];
        }
    }
    
    int ans;
    
    while (q--) {
        scanf("%d %d", &l, &r);
        l = getId(l, true);
        r = getId(r, false);
        ans = calc(l, r);
        printf("%d
", ans);
    }
}

int main() {
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif
    
    while (scanf("%d %d", &n, &q)!=EOF) {
        rep(i, 0, n)
            scanf("%d %d", &nd[i].l, &nd[i].r);
        solve();
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}

数据发生器。

from copy import deepcopy
from random import randint, shuffle
import shutil
import string


def GenDataIn():
    with open("data.in", "w") as fout:
        t = 10
        bound = 10**5
        ld = list(string.digits)
        # fout.write("%d
" % (t))
        for tt in xrange(t):
            n = randint(40, 50)
            q = randint(20, 30)
            fout.write("%d %d
" % (n, q))
            for i in xrange(n+q):
                l = randint(1, bound-1)
                r = randint(l, bound)
                fout.write("%d %d
" % (l, r))
                
            
                
def MovDataIn():
    desFileName = "F:eclipse_prjworkspacehdojdata.in"
    shutil.copyfile("data.in", desFileName)

    
if __name__ == "__main__":
    GenDataIn()
    MovDataIn()