BZOJ5142: [Usaco2017 Dec]Haybale Feast（双指针&set）（可线段树优化）

5142: [Usaco2017 Dec]Haybale Feast

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 182  Solved: 131
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Description

Farmer John is preparing a delicious meal for his cows! In his barn, he has NN haybales (1≤N≤100,0

00). The iith haybale has a certain flavor Fi (1≤Fi≤10^9) and a certain spiciness Si(1≤Si≤10^9).

The meal will consist of a single course, being a contiguous interval containing one or more consecu

tive haybales (Farmer John cannot change the order of the haybales). The total flavor of the meal is

 the sum of the flavors in the interval. The spiciness of the meal is the maximum spiciness of all h

aybales in the interval.Farmer John would like to determine the minimum spiciness his single-course 

meal could achieve, given that it must have a total flavor of at least MM (1≤M≤10^18).

给长度为n<=1e5的两个序列f[], s[] <= 1e9和一个long long M。

如果[1, n] 的一个子区间[a, b]满足 f[a]+f[a+1]+..+f[b] >= M， 我们就称[a, b]合法

，一个合法区间[a, b]的值为max(s[a], s[a+1], ..., s[b])。

让你求出可能的合法区间最小值为多少。

Input

The first line contains the integers N and M, 

the number of haybales and the minimum total flavor the meal must have, respectively. 

The next N lines describe the N haybales with two integers per line, 

first the flavor F and then the spiciness S.

Output

Please output the minimum spiciness in a single course meal that satisfies the minimum flavor requirement. 

There will always be at least one single-course meal that satisfies the flavor requirement.

Sample Input

5 10
4 10
6 15
3 5
4 9
3 6

Sample Output

9

HINT

Source

Gold

 

思路：双指针，每次得到以i为结尾的最短的大于等于M的区间的最小值，更新答案即可。

（是一个水题，但是还有更优的解法：由于最小化最大值，我们可以想到二分或者排序， 我们按照b排序，然后求最小的b，满足a的最大连续区间和大于等于M，这里可以用线段树解决。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=100010;
int a[maxn],b[maxn];
multiset<int>s;
int main()
{
    int N,head=1,ans=0; ll M,res=0;
    scanf("%d%lld",&N,&M);
    rep(i,1,N) scanf("%d%d",&a[i],&b[i]),ans=max(ans,b[i]);
    rep(i,1,N){
        res+=a[i]; s.insert(b[i]);
        while(res-a[head]>=M){
            res-=a[head]; s.erase(s.find(b[head])); head++;
        }
        if(res>=M) ans=min(ans,(*(--s.end())));
    }
    printf("%d\n",ans);
    return 0;
}