Gym 101201H Paint (离散化+DP)

题意：给定 n 个区间，让你选出一些，使得每个选出区间不交叉，并且覆盖区间最大。

析：最容易想到的先是离散化，然后最先想到的就是 O（n^2）的复杂度，dp[i] = max(dp[j] + a[i].r - a[i].l) 区间不相交，这个可以用线段树来维护一个最大值，因为有区间性，但是也可以不用线段树，直接进行线性DP，因为要选的区间越多越好，相比而言，有就要选，但是有更优的就选最优的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 50;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Node{
  LL l, r;
  bool operator < (const Node &p) const{
    return l < p.l || l == p.l && r < p.r;
  }
};
map<LL, int> mp;
Node a[maxn];
vector<LL> v;
LL dp[maxn<<1];

int main(){
  LL n;
  scanf("%I64d %d", &n, &m);
  for(int i = 0; i < m; ++i){
    scanf("%I64d %I64d", &a[i].l, &a[i].r);
    ++a[i].r;
    v.pb(a[i].l);  v.pb(a[i].r);
  }

  sort(a, a + m);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for(int i = 0; i < v.sz; ++i)  mp[v[i]] = i;
  int idx = 0;
  LL ans = 0;

  for(int i = 0; i < v.sz; ++i){
    dp[i+1] = max(dp[i+1], dp[i]);
    while(idx < m && mp[a[idx].l] == i){
      int r = mp[a[idx].r];
      dp[r] = max(dp[r], dp[i] + a[idx].r - a[idx].l);
      ++idx;
    }
    ans = max(ans, dp[i]);
  }

  printf("%I64d
", n - ans);
  return 0;
}