poj-1201 intervals (differential constraint)

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Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24379		Accepted: 9274

Description

You are given n closed, integer intervals [AI, bi] and n integers c1, ..., CN.
Write a program:
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 have at least CI common elements with interval [AI, bi], for each i= ,..., N,
Writes the answer to the standard output.

Input

The 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 & lt;= bi <= 50000 and 1 <= ci <= bi-ai+1.

Output

The 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



Test instructions

Give an interval [Ai,bi], which has the same CI elements, asking how many elements the set contains at least;


Ideas:

Can get such a few inequalities;
c[i]<=dis[b+1]-dis[a];
0<=dis[i+1]-dis[i];
-1<=dis[i]-dis[i+1];
Test instructions is must meet these conditions, then is to find the longest road, of course, can also become >= to build a map, the shortest way;



AC Code:

code for//longest path
//#include <bits/stdc++.h>#include <iostream>#include<queue>#include<cmath>#include<cstring>#include<algorithm>#include<cstdio>using namespacestd;#defineRiep (n) for (int i=1;i<=n;i++)#defineRIOP (n) for (int i=0;i<n;i++)#defineRJEP (n) for (int j=1;j<=n;j++)#defineRJOP (n) for (int j=0;j<n;j++)#defineMST (SS,B) memset (ss,b,sizeof (ss));typedefLong LongLL;ConstLL mod=1e9+7;Const DoublePi=acos (-1.0);Const intinf=0x3f3f3f3f;Const intn=5e4+5;intn,head[3*n],vis[n],ans[n],cnt,mmin;structedge{intTo,next,dis;} edge[3*N];queue<int>Qu;voidSPFA () {MST (Vis,0); MST (ans,-inf); Ans[mmin]=0;    Qu.push (mmin); Vis[mmin]=1;  while(!Qu.empty ()) {        intFr=Qu.front ();        Qu.pop ();  for(inti=head[fr];i!=-1; i=Edge[i].next) {            intx=edge[i].to; if(ans[x]<ans[fr]+Edge[i].dis) {Ans[x]=ans[fr]+Edge[i].dis; if(!Vis[x])                    {Qu.push (x); VIS[X]=1; }}} Vis[fr]=0; }}voidAdd_edge (intSintEintval) {edge[cnt].to=e; Edge[cnt].next=Head[s]; Edge[cnt].dis=Val; Head[s]=cnt++;}intMain () { while(SCANF ("%d", &n)! =EOF) {        inta,b,c,mmax=0; Mmin=55000; CNT=0; MST (Head,-1); Riep (n) {scanf ("%d%d%d",&a,&b,&c); Mmax=Max (MMAX,B); Mmin=min (mmin,a); Add_edge (A, B+1, c); } riep (Mmax) Add_edge (i,i+1,0), Add_edge (i+1, i,-1);        SPFA (); printf ("%d\n", ans[mmax+1]); }    return 0;}

//code for Shortest path//#include <bits/stdc++.h>#include <iostream>#include<queue>#include<cmath>#include<cstring>#include<algorithm>#include<cstdio>using namespacestd;#defineRiep (n) for (int i=1;i<=n;i++)#defineRIOP (n) for (int i=0;i<n;i++)#defineRJEP (n) for (int j=1;j<=n;j++)#defineRJOP (n) for (int j=0;j<n;j++)#defineMST (SS,B) memset (ss,b,sizeof (ss));typedefLong LongLL;ConstLL mod=1e9+7;Const DoublePi=acos (-1.0);Const intinf=0x3f3f3f3f;Const intn=5e4+5;intn,head[3*N],vis[n],ans[n],cnt,mmin,mmax;structedge{intTo,next,dis;} edge[3*N];queue<int>Qu;voidSPFA () {MST (Vis,0);    MST (Ans,inf); Ans[mmax]=0;    Qu.push (Mmax); Vis[mmax]=1;  while(!Qu.empty ()) {        intFr=Qu.front ();        Qu.pop ();  for(inti=head[fr];i!=-1; i=Edge[i].next) {            intx=edge[i].to; if(ans[x]>ans[fr]+Edge[i].dis) {Ans[x]=ans[fr]+Edge[i].dis; if(!Vis[x])                    {Qu.push (x); VIS[X]=1; }}} Vis[fr]=0; }}voidAdd_edge (intSintEintval) {edge[cnt].to=e; Edge[cnt].next=Head[s]; Edge[cnt].dis=Val; Head[s]=cnt++;}intMain () { while(SCANF ("%d", &n)! =EOF) {        intA,b,c; Mmax=0; Mmin=55000; CNT=0; MST (Head,-1); Riep (n) {scanf ("%d%d%d",&a,&b,&c); Mmax=max (mmax,b+1); Mmin=min (mmin,a); Add_edge (b+1, a,-c); } riep (Mmax) Add_edge (i,i+1,1), Add_edge (i+1I0);        SPFA (); printf ("%d\n",-Ans[mmin]); }    return 0;}

poj-1201 intervals (differential constraint)