zoj4028 LIS，差分约束

Posted 2020-11-02 walfy

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\n",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\n",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;
}
/***********************

***********************/

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