题意：给你以i为结尾的最长上升子序列的值，和每个值的区间范围求可行的a【i】
题解：差分约束，首先满足l[i]<=a[i]<=r[i],可以建一个虚拟节点n+1，那么有a[n+1]-a[i]<=-l[i],a[i]-a[n+1]<=r[i],同时对于之前出现过f【i】（假设为j）的情况，此时a[i]>=a[j](保证没法转移),a[j]-a[i]<=0,还有对于从上一个f[i]-1转移过来的点j,有a[i]>a[j]，即a[j]-a[i]<=-1
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define vi vector<int> #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pli pair<ll,int> #define pii pair<int,int> #define cd complex<double> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double g=10.0,eps=1e-12; const int N=100000+10,maxn=1000000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; struct edge { int to,Next,c; }e[maxn]; int l[N],r[N],f[N]; int last[N]; ll dis[N]; bool vis[N]; int head[N],cnt,n; void add(int u,int v,int c) { // printf("%d %d %d ",u,v,c); e[cnt].to=v; e[cnt].c=c; e[cnt].Next=head[u]; head[u]=cnt++; } void spfa() { memset(vis,0,sizeof vis); for(int i=1;i<=n+1;i++)dis[i]=1e18; queue<int>q; q.push(n+1); vis[n+1]=1;dis[n+1]=0; while(!q.empty()) { // printf("%d ",q.front()); int u=q.front(); q.pop(); vis[u]=0; for(int i=head[u];~i;i=e[i].Next) { int To=e[i].to; if(dis[To]>dis[u]+e[i].c) { dis[To]=dis[u]+e[i].c; if(!vis[To]) { vis[To]=1; q.push(To); } } } } bool ok=1; for(int i=1;i<=n;i++) { if(ok)printf("%d",dis[i]); else printf(" %d",dis[i]); ok=0; } puts(""); } int main() { int T;scanf("%d",&T); while(T--) { memset(last,0,sizeof last); cnt=0; memset(head,-1,sizeof head); scanf("%d",&n); for(int i=1;i<=n;i++)scanf("%d",&f[i]); for(int i=1;i<=n;i++)scanf("%d%d",&l[i],&r[i]); for(int i=1;i<=n;i++) { add(i,n+1,-l[i]); add(n+1,i,r[i]); if(last[f[i]])add(last[f[i]],i,0); if(f[i]>0)add(i,last[f[i]-1],-1); last[f[i]]=i; } spfa(); } return 0; } /*********************** ***********************/