Luogu1868 饥饿的奶牛 (动态规划)
2024-10-20 21:04:25
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;
}
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