poj 1201 Intervals 解题报告

Intervals

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

Submit Status

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

——————————————————我是华丽丽的分割线——————————————————

差分约束系统题目。

约束条件如下：

设S[i]为集合Z中小于等于i的元素个数，即S[i]=|{s|s属于Z,s<=i}|。则：

(1) S[b(i)]-S[a(i-1)]>=c(i)   ------------> S[a(i-1)]-S[b(i)]<=-c(i)

(2) S[i]-S[i-1]<=1。

(3) S[i]-S[i-1]>=0，即S[i-1]-S[i]<=0.

设所有区间极右端点为mx，极左端点为mn 则 ans=min(s[mx]-s[mn-1]) 即求源点s[mx]与s[mn-1]之间的最短路 此题得解。

改进后方法如下： (1)先只用条件(1)构图，各顶点到源点的最短距离初始为0。 (2)即刻用Bellman-ford算法求各顶点到源点的最短路径。 p.s.在每次循环中，条件1判完后判断2和3.

2的判断： s[i]<=s[i-1]+1等价于s[i]-s[mx]<=s[i-1]-s[mx]+1 设dist[i]为源点mx到i的最短路径，则s[i]-s[mx]即为dist[i]，s[i-1]-s[mx]+1即为dist[i-1]+1。 即若顶点i到源点的最短路径长度大于i-1到源点的最短路径长度+1，则dist[i]=dist[i-1]+1。

3的判断： s[i-1]<=s[i]等价于s[i-1]-s[mx]<=s[i]-s[mx] s[i]-s[mx]即为dist[i]，s[i-1]-s[mx]即为dist[i-1]。 即若顶点i-1到源点的最短路径长度大于i到源点的最短路径长度，则dist[i-1]=dist[i]。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #include<cstdlib>
 #include<iomanip>
 #include<cassert>
 #include<climits>
 #include<vector>
 #include<list>
 #include<map>
 #define maxn 1000001
 #define F(i,j,k) for(int i=j;i<=k;i++)
 #define M(a,b) memset(a,b,sizeof(a))
 #define FF(i,j,k) for(int i=j;i>=k;i--)
 #define inf 0x7fffffff
 #define maxm 2016
 #define mod 1000000007
 //#define LOCAL
 using namespace std;
 int read(){
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n,m;
 int lsh=inf,rsh;
 struct EDGE
 {
     int from;
     int to;
     int value;
     int next;
 }e[maxn];
 int head[maxn];
 int tot;
 inline void addedge(int u,int v,int w)
 {
     tot++;
     e[tot].from=u;
     e[tot].to=v;
     e[tot].value=w;
     e[tot].next=head[u];
     head[u]=tot;
 }
 int f[maxn];
 queue<int> q;
 bool vis[maxn];
 int d[maxn],in[maxn];
 bool ford()
 {
      int i,t;
      bool flag=true;
      while(flag){
          flag=false;
          F(i,,n){
              t=d[e[i].from]+e[i].value;
              if(d[e[i].to]>t){
                  d[e[i].to]=t;
                  flag=true;
              }
          }
          F(i,lsh,rsh){
              t=d[i-]+;
              if(d[i]>t){
                  d[i]=t;
                  flag=true;
              }
          }
          FF(i,rsh,lsh){
              t=d[i];
              if(d[i-]>t){
                  d[i-]=t;
                  flag=true;
              }
          }
      }
      return true;
 }
 int main()
 {
     std::ios::sync_with_stdio(false);//cout<<setiosflags(ios::fixed)<<setprecision(1)<<y;
     #ifdef LOCAL
     freopen("data.in","r",stdin);
     freopen("data.out","w",stdout);
     #endif
     cin>>n;M(head,-);
     F(i,,n){
         int a,b,c;
         cin>>a>>b>>c;
         addedge(b,a-,-c);
         if(lsh>a) lsh=a;
         if(rsh<b) rsh=b;
     }
     ford();
     cout<<d[rsh]-d[lsh-]<<endl;
     return ;
 }
 /*
 5
 3 7 3
 8 10 3
 6 8 1
 1 3 1
 10 11 1
 */

poj 1201