POJ 1201 Intervals
Intervals
Time Limit: 2000MS Memory Limit: 65536K
Total 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
5
3 7 3
8 10 3
6 8 1
1 3 1
10 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
[cpp]
#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]);
}
补充：软件开发 , C++ ,
