K. Road Widening

Posted 2020-11-27 iss-ue

(考虑每个区域可行的区间)

(x[1]=s[1] y[1]=s[1]+g[1])

(x[i]=max(x[i-1]-1,s[i]),y[i]=min(y[i-1]+1,s[i]+g[i]))

(然后这样能确保每一个区间都满足前面的区间，但不意味前面的区间满足后面的区间)

(倒过来也做一次,之后对每个都取y[i])

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2000005;
ll x[maxn],y[maxn],s[maxn],g[maxn],flag=1;
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)	cin>>s[i]>>g[i];
	x[1]=s[1],y[1]=s[1]+g[1];
	for(int i=2;i<=n;i++)
	{
		x[i]=max(x[i-1]-1,s[i]);
		y[i]=min(y[i-1]+1,s[i]+g[i]);
		if(x[i]>y[i])	flag=0;
	}
	for(int i=n-1;i>=1;i--)
	{
		x[i]=max(x[i+1]-1,x[i]);
		y[i]=min(y[i+1]+1,y[i]);
		if(x[i]>y[i])	flag=0;
	}
	ll ans=0;
	for(int i=1;i<=n;i++)	ans+=y[i]-s[i];
	if(flag==0)	cout<<-1;
	else
	{
		cout<<ans<<endl;
		for(int i=1;i<=n;i++)	cout<<y[i]<<" ";
	}
}