「CF1101F」Trucks and Cities【决策单调优化dp】

F. Trucks and Cities

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

There are $n$cities along the road, which can be represented as a straight line. The $i$-th city is situated at the distance of $a_i$ kilometers from the origin. All cities are situated in the same direction from the origin. There are ? trucks travelling from one city to another.

Each truck can be described by 4 integers: starting city $s_i$, finishing city $f_i$, fuel consumption $c_i$ and number of possible refuelings $r_i$. The $i$-th truck will spend $c_i$ litres of fuel per one kilometer.

When a truck arrives in some city, it can be refueled (but refueling is impossible in the middle of nowhere). The $i$-th truck can be refueled at most $r_i$ times. Each refueling makes truck’s gas-tank full. All trucks start with full gas-tank.

All trucks will have gas-tanks of the same size $V$ litres. You should find minimum possible $V$ such that all trucks can reach their destinations without refueling more times than allowed.

Input

First line contains two integers $n$ and $m$ $(2\le n\le 400,1\le m\le 250000)$ — the number of cities and trucks.

The second line contains $n$ integers $a_1,a_2,\dots ,a_n(1\le a_i\le 10^9,a_i<a_i+1)$ — positions of cities in the ascending order.

Next $m$ lines contains $4$ integers each. The $i$-th line contains integers $s_i,f_i,c_i,r_i(1\le s_i<f_i\le n,1\le c_i\le 10^9,0\le r_i\le n)$ — the description of the $i$-th truck.

Output

Print the only integer — minimum possible size of gas-tanks $V$ such that all trucks can reach their destinations.

Example

input

7 6
2 5 7 10 14 15 17
1 3 10 0
1 7 12 7
4 5 13 3
4 7 10 1
4 7 10 1
1 5 11 2

output

55

Note

Let’s look at queries in details:

the 1-st truck must arrive at position 7 from 2 without refuelling, so it needs gas-tank of volume at least 50.
the 2-nd truck must arrive at position 17 from 2 and can be refueled at any city (if it is on the path between starting point and ending point), so it needs gas-tank of volume at least 48.
the 3-rd truck must arrive at position 14 from 10, there is no city between, so it needs gas-tank of volume at least 52.
the 4-th truck must arrive at position 17 from 10 and can be refueled only one time: it’s optimal to refuel at 5-th city (position 14) so it needs gas-tank of volume at least 40.
the 5-th truck has the same description, so it also needs gas-tank of volume at least 40.
the 6-th truck must arrive at position 14 from 2 and can be refueled two times: first time in city 2 or 3 and second time in city 4 so it needs gas-tank of volume at least 55.

• 题意

  • 就是$x$轴上给定$n$个点「$n$范围$400$，点位置范围$10^9$」$2\times 10^5$次询问，表示询问从点a到b之间的所有线段整合成k个连续段的最大连续段长度最小

• 题解

  • 很容易想到的是直接二分找出这个最小连续段，复杂度$O(n\times m\times \log 10^9)$我也这样写了一发，虽然知道稳$T$，不过还能过43组数据

  •   #include
    
      using namespace std;
      const int maxn=250010;
      typedef long long ll;
      
      int ans[maxn],n,m,a[405],b[maxn][4];
      
      bool check(int cur,int k)
      {
      	int last=b[cur][0],cnt=0;
      	int s=b[cur][0];
      	while(s<b[cur][1]){
      		while(s+1<=b[cur][1]&&a[s+1]-a[last]<=k){
      			s++;
      		}
      		if(last==s) return false;
      		last=s;
      		cnt++;
      	}
      	return cnt-1<=b[cur][3];
      }
      
      int main()
      {
      	scanf("%d %d",&n,&m);
      	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
      	for(int i=1;i<=m;i++) {
      		scanf("%d %d %d %d",&b[i][0],&b[i][1],&b[i][2],&b[i][3]);
      		int l=1,r=a[b[i][1]]-a[b[i][0]]+1;
      		while(l<r){
      			int mid=(l+r)>>1;
      			if(check(i,mid)) r=mid;
      			else l=mid+1;
      		}
      		ans[i]=r;
      	}
      	ll res=0;
      	for(int i=1;i<=m;i++) res=max(res,(ll)b[i][2]*ans[i]);
      	printf("%lld\n",res);
      
      }
    

  • 然后考虑用$dp$解决吧，先来看看一个小一点的问题，怎么把一段区间分成若干个连续区间，最大区间长度最小。不妨定义$dp[i][j]$表示把前i个点对应的所有线段组合成j段连续线段的最大长度最小，然后转移的时候，对于$i$起那面的某个点，如果最优决策中最后一段连续区间的起点在$j$处，那么前$j$个点对应的线段一定是要划分成最优的，也就是前j段组合成$k-1$段的最大连续线段长度最小，所以就可以写转移方程式了：$$dp[i][k]=min(dp[i][k],max(dp[j][k-1],a[i]-a[j]))$$
    显然这个复杂度是$O(n^3)$，而且这个求的是线段前缀的答案，如果是所有区间的答案，则是$O(n^4)$ 对于$n=400$的数据是跑不过去的，出题人就是为了卡你这个做法范围放到了400
  • 接着考虑怎么优化这个$dp$吧，对于同一个k(分成的段数)，假设需要求的两个区间$[l,r_1]$，$[l,r_2](r_1<r_2)$那么第二个区间的最优决策对应的最后一个区间左端点$j_2$一定大于等于$j_1$，显然根据感性思想这是对的，具体严格的证明还在想。。。先假设这是对的吧，然后就好办了，对于同一个$k$，每次不用都从$1$开始找，因为决策具有单调性，用另外一个指针$point$就行了，降维到$O(n^3)$

• 代码

    #include
  
    using namespace std;
    typedef long long ll;
    
    int n,m,b,c,d,e,dp[405][405][405],a[405];
    
    int main()
    {
    	scanf("%d %d",&n,&m);
    	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    	for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++) dp[i][j][0]=a[j]-a[i];
    	for(int k=1;k<=n;k++){
    		for(int i=1;i<=n;i++){
    			int point = i;
    			for(int j=i+1;j<=n;j++){
    				while(point<j&&max(dp[i][point][k-1],a[j]-a[point])>max(dp[i][point+1][k-1],a[j]-a[point+1])) ++point;
    				dp[i][j][k]=max(dp[i][point][k-1],a[j]-a[point]);
    			}
    		}
    	}
    	ll ans=0;
    	for(int i=1;i<=m;i++) {
    		scanf("%d %d %d %d",&b,&c,&d,&e);
    		ans=max(ans,1ll*dp[b][c][e]*d);
    	}
    	printf("%lld\n",ans);
    }