Problem Statement
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is xi meters. Also, the i-th sushi has a nutritive value of vi kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
- 1≤N≤105
- 2≤C≤1014
- 1≤x1<x2<…<xN<C
- 1≤vi≤109
- All values in input are integers.
Subscores
- 300 points will be awarded for passing the test set satisfying N≤100.
Input
Input is given from Standard Input in the following format:
N C x1 v1 x2 v2 : xN vN
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Sample Input 1
3 20 2 80 9 120 16 1
Sample Output 1
191
There are three sushi on the counter with a circumference of 20 meters. If he walks two meters clockwise from the initial place, he can eat a sushi of 80 kilocalories. If he walks seven more meters clockwise, he can eat a sushi of 120 kilocalories. If he leaves now, the total nutrition taken in is 200 kilocalories, and the total energy consumed is 9 kilocalories, thus he can take in 191 kilocalories on balance, which is the largest possible value.
Sample Input 2
3 20 2 80 9 1 16 120
Sample Output 2
192
The second and third sushi have been swapped. Again, if he walks two meters clockwise from the initial place, he can eat a sushi of 80 kilocalories. If he walks six more meters counterclockwise this time, he can eat a sushi of 120 kilocalories. If he leaves now, the total nutrition taken in is 200 kilocalories, and the total energy consumed is 8 kilocalories, thus he can take in 192 kilocalories on balance, which is the largest possible value.
Sample Input 3
1 100000000000000 50000000000000 1
Sample Output 3
0
Even though the only sushi is so far that it does not fit into a 32-bit integer, its nutritive value is low, thus he should immediately leave without doing anything.
Sample Input 4
15 10000000000 400000000 1000000000 800000000 1000000000 1900000000 1000000000 2400000000 1000000000 2900000000 1000000000 3300000000 1000000000 3700000000 1000000000 3800000000 1000000000 4000000000 1000000000 4100000000 1000000000 5200000000 1000000000 6600000000 1000000000 8000000000 1000000000 9300000000 1000000000 9700000000 1000000000
Sample Output 4
6500000000
All these sample inputs above are included in the test set for the partial score.
meaning of the title
n lines, each line x[i] and v[i] represent the position of the sushi and the energy obtained after eating
There is a circular counter in the restaurant with a perimeter of C. Nakahashi starts from 1 and consumes 1k calories for every 1m he walks. Nakahashi can leave the restaurant at any time and ask for the maximum energy that Nakahashi can take away.
answer
Considering n<=1e5, shun[i] goes to i along the way, and ni[i] goes to i in the opposite direction
There is a double cycle of forward and reverse i, j is judged to follow to i, and reverse to the maximum of j
Or only the straight line, or only the reverse, the maximum of the three cases can only get 300 points
How to optimize the complexity?
We can think about it in this way, shun[i] goes to the maximum of i along the way, and ni[i] goes to the maximum of i in the opposite direction
If there is an ups and downs, consider going down to i first, and then down to i+1
Then consider walking backwards to i, and walking to i-1
Either only the straight, or only the inverse
code
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int N=1e5+5; 4 typedef long long ll; 5 ll n,c,x[N],w[N],val[N],rval[N],shun[N],ni[N]; 6 int main() 7 { 8 cin>>n>>c; 9 for(int i=1;i<=n;i++) 10 { 11 cin>>x[i]>>w[i]; 12 val[i]=val[i-1]+w[i];//Follow the value of i to the sum of 13 shun[i]=max(shun[i- 1 ],val[i]-x[i]); // Follow to i to take the maximum 14 } 15 for ( int i=n ;i>= 1 ;i-- ) 16 { 17 rval[i]=rval[i+ 1 ]+w[i]; // reverse to the sum of the value of i 18 ni[i]=max(ni[i+ 1 ],rval[i]-c+x[i]); // Reverse to i and take the maximum 19 } 20 ll maxx= 0 ; 21 for ( int i=n;i>= 1 ;i-- ) 22 { 23 maxx=max(maxx,rval[i]- 2 *(cx[i])+shun[i- 1 ]); // First go back to i, then go to i-1 24 maxx=max( maxx,ni[i]); // only inverse 25 } 26 for ( int i= 1 ;i<=n;i++ ) 27 { 28 maxx=max(maxx,val[i]- 2 *x[i]+ ni[i+ 1 ]); // Follow to i first, then reverse to i+1 29 maxx=max(maxx,shun[i]); // Only follow 30 } 31 cout<< maxx; 32 return 0; 33 }