POJ1201（Hdu1384） Intervals差分约束系统

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 endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi - ai + 1.

Process to the end of file.

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

题意大概是这样：有n个区间[Ai，Bi],有一个集合Z，集合Z和每个区间至少有Ci个相同元素；求这个集合至少有多少个元素。

定义dis
数组记录区间[0,n]的最少有多少个元素在集合z内，则可以建立一个不等式：dis[Bi+1]-dis[Ai]>=Ci；

另外由题意有：0<=dis[i+1]-dis[i]<=1;（因为i可能从0开始所以不使用i-1和i）

因此有：

dis[i+1]-dis[i]>=0；

dis[i]-dis[i+1]>=-1；

因为是求最小值，所以是计算最长路。

边数比较多而且有负边所以使用spfa算法。

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define INF 1<<29
using namespace std;
struct node
{
int to,len,next;
}edge[200050];
int head[50010],used[50010],dis[50010],inqueue[50010],k,n,minx,maxx;
void add_edge(int st,int en,int len)
{
edge[k].to=en;
edge[k].len=len;
edge[k].next=head[st];
head[st]=k++;
}
void init()
{
memset(used,0,sizeof(used));
memset(inqueue,0,sizeof(inqueue));
memset(head,-1,sizeof(head));
minx=INF,maxx=-1;
for(int i=1;i<=n;i++)
{
int st,en,len;
scanf("%d%d%d",&st,&en,&len);
minx=min(minx,st);
maxx=max(maxx,en+1);
add_edge(st,en+1,len);
}
for(int i=minx;i<=maxx;i++)
{
add_edge(i,i+1,0);
add_edge(i+1,i,-1);
}
}
void spfa(int st,int time)
{
for(int i=minx;i<=maxx;i++)
dis[i]=-INF;
queue<int>q;
dis[st]=0;
used[st]=1;
q.push(st);
inqueue[st]++;
while(!q.empty())
{
int top=q.front();
q.pop();
used[top]=0;
for(int j=head[top];j!=-1;j=edge[j].next)
{
if(dis[edge[j].to]<dis[top]+edge[j].len)
{
dis[edge[j].to]=dis[top]+edge[j].len;
if(!used[edge[j].to])
{
used[edge[j].to]=1;
if(++inqueue[edge[j].to]>time-1)
{
printf("%d\n",dis[maxx]);
}
q.push(edge[j].to);
}
}
}
}
printf("%d\n",dis[maxx]);
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
k=0;
init();
spfa(minx,maxx-minx+1);
}
return 0;
}