Luogu1868 饥饿的奶牛 （动态规划）

开始以为是贪心，10分；想了个DP估计会超时，一翻题解各路初中神仙，背包都有。

\(n^2\)很好想，考虑单调性用二分优化出log

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long
#define ON_DEBUG
#ifdef ON_DEBUG
#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);
#else
#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;
#endif
struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;
const int N = 150007;
struct Field{
	int l, r, w;
	bool operator < (const Field &com) const{
		if(r != com.r) return r < com.r;
		return l < com.l;
	}
}a[N];
int f[N];
inline int Find(int x){
	int l = 1, r = x;
	int ans = -1;
	while(l <= r){
		int mid = (l + r) >> 1;
		if(a[mid].r < a[x].l){
			l = mid + 1;
			ans = mid;
		}
		else{
			r = mid - 1;
		}
	}
	return ans;
}
int main(){
	//FileOpen();
	int n;
	io >> n;
    R(i,1,n){
    	io >> a[i].l >> a[i].r;
    	a[i].w = a[i].r - a[i].l + 1;
    }
    sort(a + 1, a + n + 1);
    f[1] = a[1].r - a[1].l + 1;
    R(i,2,n){
        int j = Find(i);
        if(j != -1)
        	f[i] = Max(f[i - 1], f[j] + a[i].w);
        else
        	f[i] = Max(f[i - 1], a[i].w);
    }
    printf("%d\n",f[n]);
    return 0;
}