（差分约束） hdu 1384

Intervals

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3038 Accepted Submission(s): 1118


[align=left]Problem Description[/align]
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

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1

[align=left]Sample Output[/align]

6

[align=left]Author[/align]
1384

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0xfffffff
vector<int> e[50001],w[50001];
queue<int> q;
int n,dist[50001],r,l,vis[50001];
void addedge(int a,int b,int c)
{
e[b].push_back(a);
w[b].push_back(c);
}
void SPFA()
{
int x;
vis[r]=1;
q.push(r);
while(!q.empty())
{
x=q.front(),q.pop();
vis[x]=0;
for(int i=0;i<e[x].size();i++)
{
if(dist[x]+w[x][i]<dist[e[x][i]])
{
dist[e[x][i]]=dist[x]+w[x][i];
if(!vis[e[x][i]])
{
vis[e[x][i]]=1;
q.push(e[x][i]);
}
}
}
}
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
int a,b,c;
l=INF,r=0;
for(int i=0;i<50001;i++)
{
vis[i]=0;
e[i].clear(),w[i].clear();
dist[i]=INF;
}
for(int i=0;i<n;i++)
{
scanf("%d%d%d",&a,&b,&c);
addedge(a-1,b,-c);
if(a<l) l=a;
if(b>r) r=b;
}
l--;
for(int i=l;i<r;i++)
{
addedge(i+1,i,1);
addedge(i,i+1,0);
}
dist[r]=0;
while(!q.empty()) q.pop();
SPFA();
printf("%d\n",-dist[l]);
}
return 0;
}