POJ1201 Intervals

 

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 25125		Accepted: 9580

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

差分约束

/**/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int mxn=60000;
struct edge{
    int v,dis;
    int next;
}e[mxn*10];
int hd[mxn],cnt;
int dis[mxn];
int vis[mxn];
int inq[mxn];
int n;
void add_edge(int u,int v,int w){
    e[++cnt].next=hd[u];e[cnt].v=v;e[cnt].dis=w;hd[u]=cnt;
}
bool SPFA(){
    memset(dis,-1,sizeof dis);
    queue<int>q;
    q.push(0);
    inq[0]=1;
    vis[0]=1;
    dis[0]=0;
    while(!q.empty()){
        int u=q.front();
        for(int i=hd[u];i;i=e[i].next){
            int v=e[i].v;
            if(dis[u]+e[i].dis>dis[v]){
                dis[v]=dis[u]+e[i].dis;
                vis[v]++;
                if(vis[v]>n)return 0;
                if(!inq[v]){
                    inq[v]=1;
                    q.push(v);
                }
            }
        }
        q.pop();
        inq[u]=0;
    }
    return 1;
}
int main(){
    scanf("%d",&n);
    int i,j;
    int mx=-1,a,b,c;
    for(i=1;i<=n;i++){
        scanf("%d%d%d",&a,&b,&c);
        add_edge(a,b+1,c);
        mx=max(mx,b+1);
    }
    for(i=0;i<mx;i++){
        add_edge(i,i+1,0);
        add_edge(i+1,i,-1);
    }
    SPFA();
    printf("%d
",dis[mx]);
    return 0;
}