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;
}