图论--差分约束--POJ 1201 Intervals

Intervals

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 30971 Accepted: 11990

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

这个题我看了很久都不懂得怎么建图。先粘个代码吧！

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define inf 99999
#define maxx 50005
struct edge
{
    int u,v,w;
}edge[maxx];
int n;
int dist[maxx];
int l;//左端点的最小值
int r;//右端点的最大值

void bellman_ford()
{
    int flag=1,t;
    while(flag)
    {
        flag=0;
        for(int i=1;i<=n;i++)
        {
            t=dist[edge[i].u]+edge[i].w;
            if(dist[edge[i].v]>t)
            {
                dist[edge[i].v]=t;
                flag=1;
            }
        }
        for(int i=r;i>l;i--)
        {
            t=dist[i-1]+1;
            if(dist[i]>t)
            {
                dist[i]=t;flag=1;
            }
        }
        for(int i=r;i>l;i--)
        {
            t=dist[i];
            if(dist[i-1]>t)
            {
                dist[i-1]=t;
                flag=1;
            }
        }
    }
}

int main()
{
    scanf("%d",&n);
    r=1,l=inf;
    int a,b,c;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        edge[i].u=b,edge[i].v=a-1,edge[i].w=-c;
        if(a<l)  l=a;
        if(b>r)    r=b;
        dist[i]=0;
    }
    bellman_ford();
    printf("%d\n",dist[r]-dist[l-1]);
    return 0;
}