https://atcoder.jp/contests/arc166/tasks/arc166_d

首先肯定是这样子放：

考虑相邻之间的差，本质就是橙色区间减蓝色区间数量

区间数量越少显然越优，所以我们要么保留橙区间，要么保留紫区间，然后两两匹配

#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar(); while(ch<'0'||
ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
//mt19937 rand(time(0));
//mt19937_64 rand(time(0));
//srand(time(0));
#define N 400010
//#define M
//#define mo
struct node {int op, x, y; 
}a[N], t;
queue<node>q; 
int n, m, i, j, k, T, ans;
int x[N], y[N], d; bool cmp(node x, node y) {if(x.x==y.x) return x.op<y.op; return x.x<y.x; 
}signed main()
{
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
//	T=read();
//	while(T--) {
//
//	}n=read(); x[0]=-1e12; x[n+1]=1e12; ans=1e12; for(i=1; i<=n; ++i) x[i]=read(); for(i=1; i<=n; ++i) y[i]=read(); for(i=1; i<=n+1; ++i)  {d=y[i]-y[i-1]; if(d>0) a[++k].x=x[i-1]+1, a[k].y=d, a[k].op=1; if(d<0) a[++k].x=x[i]-1, a[k].y=-d, a[k].op=2; }sort(a+1, a+k+1, cmp); for(i=1; i<=n; ++i) {
//		printf("%lld : %lld %lld\n", a[i].op, a[i].x, a[i].y); if(a[i].op==1) q.push(a[i]); else {while(!q.empty() && q.front().y<=a[i].y) {t=q.front(); q.pop(); ans=min(ans, a[i].x-t.x); a[i].y-=t.y; }if(a[i].y) {t=q.front(); q.front().y-=a[i].y; ans=min(ans, a[i].x-t.x); }}}if(ans>1e9) printf("-1"); else printf("%lld", ans); return 0;
}