poj1201 差分约束系统/system of different contraints

原创于 2011-06-01 23:10:00 发布 · 318 阅读

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CC 4.0 BY-SA版权

文章标签：

#intervals #integer #input #bi #output #each

本文介绍了一个区间问题的求解算法，通过Bellman-Ford算法计算最小集合大小，使得该集合与每个给定区间至少有ci个公共元素。适用于解决特定类型的最优化问题。

用s[i]表示小于等于i的元数个数

可得约束条件为：s[node.from]-s[node.to]<-c,隐含的为0<=s[i]-s[i-1]<=1

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=50010;
struct cnode
{
int from;
int to;
int w;
}node[MAXN];
int d[MAXN];
int n;
int mmin;
int mmax;
bool bellman_ford()
{
memset(d,0,sizeof(d));
bool flag=1;
bool f;
int i;
while(flag)
    {
      flag=0;
      for(i=0;i<n;i++)
    {
    if(d[node[i].from]>d[node[i].to]+node[i].w)//from-to<c
        {
          d[node[i].from]=d[node[i].to]+node[i].w;
          flag=1;
        }
    }


      for(i=mmin;i<=mmax;i++)
    {
      if(d[i-1]+1<d[i])//0<=d[i]-d[i-1]<=1
    {
    d[i]=d[i-1]+1;
    f=1;
    }
    }


for(i=mmax;i>=mmin;i--)
    {
      if(d[i-1]>d[i])
    {
    d[i-1]=d[i];
    f=1;
    }
    }

    }
return true;

}
int main()
{
    int FROM,TO,W;
int i;
while(scanf("%d",&n)!=EOF)
    {
      mmin=800000000;
      mmax=-100;
      for(i=0;i<n;i++)
    {
    scanf("%d %d %d",&FROM,&TO,&W);
    node[i].from=FROM-1;
    node[i].to=TO;
    node[i].w=-W;
    if(FROM<mmin)
        {
          mmin=FROM;
        }
    if(TO>mmax)
        {
          mmax=TO;
        }

    }
      bellman_ford();
      printf("%d/n",d[mmax]-d[mmin-1]);
      return 0;
    }

return 0;
}

附题目：

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13145		Accepted: 4860

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