【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+;
 int n, q;
 int dp[][maxn*];
 int X[maxn*], 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 - ;
 }

 int calc(int l, int r) {
     if (l > r)
         return ;

     if (l==nx || r==-)
         return ;

     int ret = ;
     per(i, , ) {
         if (dp[i][l] <= r) {
             ret += (<<i);
             l = dp[i][l];
         }
     }

     return ret;
 }

 void solve() {
     nx = ;
     rep(i, , 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, , n) {
         nd[i].l = getId(nd[i].l, true);
         nd[i].r = getId(nd[i].r, false);
     }

     int mn = nx, l, r = n-;
     per(i, , nx) {
         while (r>= && nd[r].l>=i) {
             mn = min(mn, nd[r].r);
             --r;
         }
         dp[][i] = mn;
     }
     rep(i, , )
         dp[i][nx] = nx;
     rep(i, , ) {
         rep(j, , nx) {
             dp[i][j] = dp[i-][dp[i-][j]];
         }
     }

     int ans;

     while (q--) {
         scanf("%d %d", &l, &r);
         l = getId(l, true);
         r = getId(r, false);
         ans = calc(l, r);
         printf("%d\n", 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, , n)
             scanf("%d %d", &nd[i].l, &nd[i].r);
         solve();
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }

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\n" % (t))
         for tt in xrange(t):
             n = randint(40, 50)
             q = randint(20, 30)
             fout.write("%d %d\n" % (n, q))
             for i in xrange(n+q):
                 l = randint(1, bound-1)
                 r = randint(l, bound)
                 fout.write("%d %d\n" % (l, r))

 def MovDataIn():
     desFileName = "F:\eclipse_prj\workspace\hdoj\data.in"
     shutil.copyfile("data.in", desFileName)

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