POJ 1201 Intervals

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18740		Accepted: 7042

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

 

 

#include <stdio.h>
#include <string.h>
#define EM 300000
#define VM 50005

struct edge
{
    int v,w,next;
}e[EM];
int head[VM],ep;

void addedge(int cu,int cv,int cw)
{
    ep ++;
    e[ep].v = cv;
    e[ep].w = cw;
    e[ep].next = head[cu];
    head[cu] = ep;
}
int maxn (int a,int b)
{
    return a > b?a:b;
}

void spfa (int n)
{
    int vis[VM],dis[VM],stack[EM];
    memset (vis,0,sizeof(vis));
    memset (dis,-1,sizeof(dis));
    //for (int i = 0;i <= n;i ++)
     //   dis[i] = inf;
    dis[n+2] = 0;
    int top = 1;
    vis[n+2] = 1;
    stack[0] = n+2;
    while (top)
    {
        int u = stack[--top];
        vis[u] = 0;
        for (int i = head[u];i != -1;i = e[i].next)
        {
            int v = e[i].v;
            if (dis[v] < dis[u] + e[i].w)
            {
                dis[v] = dis[u] + e[i].w;
                if (!vis[v])
                {
                    vis[v] = 1;
                    stack[top++] = v;
                }
            }
        }
    }
    printf ("%d\n",dis[n]);
}
int main ()
{
    int n,v1,v2,m,cost;
    ep = 0;
    scanf ("%d",&n);
    m = n;
    memset (head,-1,sizeof(head));
    int max = -1;
    while (m --)
    {
        scanf ("%d%d%d",&v1,&v2,&cost);
        addedge (v1,v2+1,cost);
        max = maxn (max,v2+1);
    }
    for (int i = 0;i < max;i ++)
    {
        addedge (i,i+1,0);
        addedge (i+1,i,-1);
        addedge (max+2,i,0);
    }
    spfa (max);
    return 0;
}