POJ 1201 Intervals

Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 28852   Accepted: 11131

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

#include <map>
#include <set>
#include <stack>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-15
#define MAXN  50005
#define INF 1000000007
#define MAX(a,b)(a>b?a:b)
#define MIN(a,b)(a<b?a:b)
int N,n,Max,Min;
int head[MAXN],d[MAXN],vis[MAXN];
struct EDG{ int v,w,next; }edge[3*MAXN];
void AddEdge(int u,int v,int w){
    edge[N].v=v;edge[N].w=w;edge[N].next=head[u];head[u]=N++;
}
void SPFA(){
    for(int i=Min;i<=Max;i++) d[i]=-INF;
    d[Min]=0;queue<int>q;q.push(Min);
    while(!q.empty()){
        int x=q.front();q.pop();
        vis[x]=0;
        for(int e=head[x];e!=-1;e=edge[e].next)
            if(d[edge[e].v]<d[x]+edge[e].w){
                d[edge[e].v]=d[x]+edge[e].w;
                if(!vis[edge[e].v]){
                    q.push(edge[e].v);
                    vis[edge[e].v]=1;
                }
            }
    }
} 
int main(){
    while(scanf("%d", &n)!=EOF){
        memset(vis,0,sizeof(vis));
        memset(edge,0,sizeof(edge));
        memset(head,-1,sizeof(head));
        Min=INF;Max=-INF;N=0;
        for(int i=0;i<n;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            AddEdge(u,v+1,w);
            Min=MIN(Min,u);
            Max=MAX(Max,v+1);
        }
        for(int i=Min;i<Max;i++){
            AddEdge(i,i+1,0);
            AddEdge(i+1,i,-1);
        }
        SPFA();
        printf("%d\n", d[Max]);
    }
} 
/*
5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1
*/

 

posted @ 2018-05-26 10:42  一蓑烟雨任生平  阅读(125)  评论(0编辑  收藏  举报