hdoj1384Intervals【差分约束】

Intervals

  Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  

  Total Submission(s): 3556    Accepted Submission(s): 1303


Problem 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

 

Author

1384

 
题意：给定多个区间[a,b]存在一个集合A在区间[a,b]内至少有c个数在集合A中求A中最少有多少个元素；
设S[i]为集合A在区间0到i中的元素个数则有题意知有在区间[a,b]对集合S中则满足S[b]-S[a-1]>=c;另有0=<S[i]-S[i-1]<=1;
既有S[a-1]-S[b]<=-c;S[i-1]-S[i]<=0;S[i]-S[i-1]<=1;

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=50010;
struct Node{
int from,to,value;
int next;
}A[maxn<<2];
int dist[maxn];
bool vis[maxn];
int head[maxn],node;
void init(){
node=0;
memset(vis,false,sizeof(vis));
memset(head,-1,sizeof(head));
memset(dist,0x3f,sizeof(dist));
}
void add(int u,int v,int value){
A[node].from=u;
A[node].to=v;
A[node].value=value;
A[node].next=head[u];
head[u]=node++;
}
int spfa(int minl,int maxr){
int i,j,k,u;
queue<int>Q;
dist[maxr]=0;
vis[maxr]=true;
Q.push(maxr);
while(!Q.empty()){
u=Q.front();Q.pop();vis[u]=false;
for(k=head[u];k!=-1;k=A[k].next){
int v=A[k].to;
if(dist[v]>dist[u]+A[k].value){
dist[v]=dist[u]+A[k].value;
if(!vis[v]){
vis[v]=true;
Q.push(v);
}
}
}
}
return -dist[minl-1];
}
int main()
{
int n,i,j,k;
while(scanf("%d",&n)!=EOF){
int a,b,c;init();
int minl=inf,maxr=0;
for(i=0;i<n;++i){
scanf("%d%d%d",&a,&b,&c);
add(b,a-1,-c);
minl=min(minl,a);
maxr=max(maxr,b);
}
for(i=minl;i<=maxr;++i){
add(i-1,i,1);
add(i,i-1,0);
}
printf("%d\n",spfa(minl,maxr));
}
return 0;
}