POJ1201 Intervals (difference constraint), poj1201intervals

POJ1201 Intervals (difference constraint), poj1201intervals

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:28416		Accepted:10966

DescriptionYou are given n closed, integer intervals [ai, bi] and n integers c1,..., cn.
Write a program that:
Reads the number of intervals, their end points and integers c1,..., cn from the standard input,
Computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each I = 1, 2,..., n,
Writes the answer to the standard output. inputThe first line of the input contains an integer n (1 <= n <= 50000) -- the number of intervals. the following n lines describe the intervals. the (I + 1)-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50000 and 1 <= ci <= bi-ai + 1. outputThe output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each I = 1, 2 ,..., n.

Sample Input

53 7 38 10 36 8 11 3 110 11 1

Sample Output

6

SourceSouthwestern Europe 2002 Each time you give an interval $ [a_ I, B _ I] $ and a number $ c_ I $, so that there is at least $ c_ I $ in the middle, and you can find a minimum set $ Z $, so that the set $ Z $ meets all the above requirements, ask the size of the set $ Z $ Ideas: Set $ S [I] $ to indicate the prefix and Then the relation of the question becomes $ S [B _ I]-S [a_ I]> = c_ I $ This is a typical differential constraint problem. The minimum set is required in the question, so convert to the longest path and write all the statements in the form of $ B-A> = C $ There is also a condition $0 <= S [I]-S [I-1] <= 1 $ Because the data is an integer So we get two more equations. $ S \ left [I \ right]-S \ left [i-1 \ right] \ geq 0 $
$ S \ left [i-1 \ right]-S \ left [I \ right] \ geq-1 $ But there is a detail: $ S [I-1] $ cannot be represented, so we need to place all subscript $ + 1 $, then $ S [I] $ represents $0 to (I-1) $ prefix and At the same time, this figure is China Unicom, so you do not need to create a super Source 

#include<cstdio>#include<queue>#include<cstring>#define INF 1e8+10using namespace std;const int MAXN=1e6+10;#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++)char buf[MAXN],*p1=buf,*p2=buf;inline int read(){    char c=getchar();int x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}    return x*f;}struct node{    int u,v,w,nxt;}edge[MAXN];int head[MAXN],num=1;int maxx=-INF,minn=INF;int dis[MAXN],vis[MAXN];inline void AddEdge(int x,int y,int z){    edge[num].u=x;    edge[num].v=y;    edge[num].w=z;    edge[num].nxt=head[x];    head[x]=num++;}int SPFA(){    queue<int>q;    memset(dis,-0xf,sizeof(dis));    dis[minn]=0;q.push(minn);     while(q.size()!=0)    {        int p=q.front();q.pop();        vis[p]=0;        for(int i=head[p];i!=-1;i=edge[i].nxt)        {            if(dis[edge[i].v]<dis[p]+edge[i].w)            {                dis[edge[i].v]=dis[p]+edge[i].w;                if(vis[edge[i].v]==0)                    vis[edge[i].v]=1,q.push(edge[i].v);            }        }    }    printf("%d",dis[maxx]);}int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    memset(head,-1,sizeof(head));    int N=read();    for(int i=1;i<=N;i++)    {        int x=read(),y=read(),z=read();        AddEdge(x,y+1,z);         maxx=max(y+1,maxx);        minn=min(x,minn);    }    for(int i=minn;i<=maxx-1;i++)    {        AddEdge(i+1,i,-1);        AddEdge(i,i+1,0);    }    SPFA();    return 0;}