Educational Codeforces Round 56 (Rated for Div. 2) C - Mishka and the Last Exam(贪心/差分约束)

题意

给一个n，一个序列b[]，

bi=ai+a(n+i-1)，

求不降序的a序列[]

思路来源

组里各神犇&&自己

题解

Solution1:

显然a1=0，an=b1的时候，区间长度最长

区间里面内置区间的时候如果内区间能左对齐，显然内区间最长，

即，如果能[0,8]，显然不要[1,7]

而若不能左对齐的话，就一定要右对齐

即，如果能[3,10]，显然不要[6,7]，右对齐更长，

那么由于保证外区间都是最长的，

如果存在解，贪心最长区间一定是最优的

Solution2:

0<=a1<=a2<=…<=an/2<=bn/2-an/2<=…<=b1-a1

令a0=0，跑一遍差分约束的最长路，求最小值即可

注意从a0出发，spfa(0)，初始化赋负INF

代码

Solution1:(贪心O(n))

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
const int maxn=2e5+10;
int n;
ll a[maxn],b[maxn/2];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n/2;++i)scanf("%I64d",&b[i]);
	a[1]=0,a[n]=b[1];
	for(int i=2;i<=n/2;++i)
	{
		int r=n-i+1;
		a[i]=a[i-1];
		a[r]=b[i]-a[i];
		if(a[r]>a[r+1])
		{
			a[r]=a[r+1];
			a[i]=b[i]-a[r];
		}
	}
	for(int i=1;i<=n;++i)
	{
		printf("%I64d%c",a[i],i==n?'\n':' ');
	}
	return 0;
}

Solution2:(差分约束O(nlogn))

By Yzm007, contest: Educational Codeforces Round 56 (Rated for Div. 2), problem: (C) Mishka and the Last Exam, Accepted, #
 #include <iostream>
#include <algorithm> 
#include <cstring>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <functional>
const int INF=0x3f3f3f3f;
const int maxn=2e5+10; 
const int mod=1e9+7;
const int MOD=998244353;
const double eps=1e-7;
typedef long long ll;
#define vi vector<int> 
#define si set<int>
#define pii pair<int,int> 
#define pi acos(-1.0)
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
#define sci(x) scanf("%d",&(x))
#define scll(x) scanf("%I64d",&(x))
#define sclf(x) scanf("%lf",&(x))
#define pri(x) printf("%d",(x))
#define rep(i,j,k) for(int i=j;i<=k;++i)
#define per(i,j,k) for(int i=j;i>=k;--i)
#define mem(a,b) memset(a,b,sizeof(a)) 
using namespace std;
int n,head[maxn],cnt;
ll b[maxn/2],a[maxn],dis[maxn],q[10*maxn];
bool vis[maxn];
struct edge
{
 int to,nex;
 ll w;
}e[maxn];
void init()
{
	cnt=0;
	mem(head,-1);
}
void spfa(int s)
{
	for(int i=0;i<maxn;++i)dis[i]=-8e18;
	mem(vis,0);
	int start=0,end=1;
	q[0]=s;dis[s]=0;
	while(start<end)
	{
		int u=q[start++];vis[u]=0;
		for(int i=head[u];~i;i=e[i].nex)
		{
			int v=e[i].to;
			ll w=e[i].w;
			if(dis[v]<dis[u]+w)
			{
				dis[v]=dis[u]+w;
				if(!vis[v])
				{
					q[end++]=v;
					vis[v]=1;
				}
			}
		}
	}
}
void add(int u,int v,ll w)
{
	e[cnt].to=v;
	e[cnt].w=w;
	e[cnt].nex=head[u];
	head[u]=cnt++;
}
int main()
{ 
   init();
   sci(n);
   rep(i,1,n/2)scll(b[i]);
   rep(i,2,n/2)
   {
   	add(i-1,i,b[i]-b[i-1]);//ai-ai-1>=k
   	add(i-1,i,0);//ai-ai-1>=0
   }
   add(0,1,0);//a1-a0>=0
   add(n/2,0,-b[n/2]/2);//a[n/2]<=b[n/2]-a[n/2]
   spfa(0);
   rep(i,1,n/2)
   {
    a[i]=dis[i];
    a[n-i+1]=b[i]-dis[i];
   }
   rep(i,1,n)
   {
   	printf("%I64d%c",a[i],i==n?'\n':' ');
   }
   return 0;
}