POJ 1201 Intervals

Intervals

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 18456 Accepted: 6919

Description

You 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.

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 <= 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

Source

Southwestern Europe 2002

     考差分约束，自我感觉是这类问题是找不等式关系，然后想方法建图，然后用一遍spfa就可以了

 初步接触差分约束

#include <stdio.h>#include <string.h>#include <math.h>struct num{    int end,val,next;}a[1000000];int b[50100];int d[50100],queue[1000000];int status[50100];int INF=0x7fffff;int main(){    int spfa(int sta,int end);    int i,j,n,m,s,t;    int x,y,val,max,min;    while(scanf("%d",&n)!=EOF)    {        memset(b,-1,sizeof(b));        max=-1; min=INF;        for(i=0,j=0;i<=n-1;i++)        {            scanf("%d %d %d",&x,&y,&val);            x++; y++;            if(x-1<min)            {                min=x-1;            }            if(y>max)            {                max=y;            }            a[j].end= y;            a[j].val=val;            a[j].next=b[x-1];            b[x-1]=j; j++;        }        for(i=min+1;i<=max;i++)        {            a[j].end=i;            a[j].val=0;            a[j].next=b[i-1];            b[i-1]=j; j++;            a[j].end=i-1;            a[j].val=-1;            a[j].next=b[i];            b[i]=j; j++;        }        t=spfa(min,max);        printf("%d\n",t);    }    return 0;}int spfa(int sta,int end){    int i,j,top,base,x,xend,y;    for(i=sta;i<=end;i++)    {        d[i]=-1*INF;    }    top=base=0;    queue[top++]=sta;    d[sta]=0;    memset(status,0,sizeof(status));    status[sta]=1;    while(base<top)    {        x=queue[base++];        status[x]=0;        for(j=b[x];j!=-1;j=a[j].next)        {            y=a[j].end;            if(d[y]<(d[x]+a[j].val))            {                d[y]=d[x]+a[j].val;                if(!status[y])                {                    queue[top++]=y;                }            }        }    }    return (d[end]);}