Solution: poi 2009 lyz

Question Surface

Hall Theorem in Board Discussion

Hall theorem: a bipartite graph has a perfect matching condition. For any $ I $ left vertex, there are at least $ I $ right vertex adjacent to them. Put it in this question, that is, it is obvious that the case that the shoes are not enough is a continuous person, so we can maintain the maximum sub-section and just like =. =

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 const int N=200005; 6 long long ll[4*N],rr[4*N],len[4*N],val[4*N]; 7 long long n,m,k,d,t1,t2; 8 void pushup(int nde) 9 {10     int ls=2*nde,rs=2*nde+1;11     ll[nde]=max(val[ls]+ll[rs],ll[ls]);12     rr[nde]=max(val[rs]+rr[ls],rr[rs]);13     len[nde]=max(max(len[ls],len[rs]),rr[ls]+ll[rs]);14     val[nde]=val[ls]+val[rs];15 }16 void create(int nde,int l,int r)17 {18     if(l==r) 19         ll[nde]=rr[nde]=len[nde]=val[nde]=-k;20     else21     {22         int mid=(l+r)/2,ls=2*nde,rs=2*nde+1;23         create(ls,l,mid),create(rs,mid+1,r);24         pushup(nde);25     }26 }27 void change(int nde,int l,int r,int pos,int task)28 {29     if(l==r) 30     {31         ll[nde]+=task,rr[nde]+=task;32         len[nde]+=task,val[nde]+=task;33     }34     else35     {36         int mid=(l+r)/2,ls=2*nde,rs=2*nde+1;37         if(pos<=mid) change(ls,l,mid,pos,task);38         else change(rs,mid+1,r,pos,task);39         pushup(nde);40     }41 }42 int main ()43 {44     scanf("%lld%lld%lld%lld",&n,&m,&k,&d),create(1,1,n);45     for(int i=1;i<=m;i++)46     {47         scanf("%lld%lld",&t1,&t2),change(1,1,n,t1,t2);48         (len[1]<=k*d)?printf("TAK\n"):printf("NIE\n");49     }50     return 0;51 }52 

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Solution: poi 2009 lyz