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

53 7 38 10 36 8 11 3 110 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]。

  1 #include<iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<cmath>  5 #include<algorithm>  6 #include<queue>  7 #include<cstdlib>  8 #include<iomanip>  9 #include<cassert> 10 #include<climits> 11 #include<vector> 12 #include<list> 13 #include<map> 14 #define maxn 1000001 15 #define F(i,j,k) for(int i=j;i<=k;i++) 16 #define M(a,b) memset(a,b,sizeof(a)) 17 #define FF(i,j,k) for(int i=j;i>=k;i--) 18 #define inf 0x7fffffff 19 #define maxm 2016 20 #define mod 1000000007 21 //#define LOCAL 22 using namespace std; 23 int read(){ 24     int x=0,f=1;char ch=getchar(); 25     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 26     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 27     return x*f; 28 } 29 int n,m; 30 int lsh=inf,rsh; 31 struct EDGE 32 { 33     int from; 34     int to; 35     int value; 36     int next; 37 }e[maxn]; 38 int head[maxn]; 39 int tot; 40 inline void addedge(int u,int v,int w) 41 { 42     tot++; 43     e[tot].from=u; 44     e[tot].to=v; 45     e[tot].value=w; 46     e[tot].next=head[u]; 47     head[u]=tot; 48 } 49 int f[maxn]; 50 queue<int> q; 51 bool vis[maxn]; 52 int d[maxn],in[maxn]; 53 bool ford()  54 { 55      int i,t; 56      bool flag=true; 57      while(flag){ 58          flag=false; 59          F(i,1,n){ 60              t=d[e[i].from]+e[i].value; 61              if(d[e[i].to]>t){ 62                  d[e[i].to]=t; 63                  flag=true; 64              } 65          } 66          F(i,lsh,rsh){ 67              t=d[i-1]+1; 68              if(d[i]>t){ 69                  d[i]=t; 70                  flag=true; 71              } 72          } 73          FF(i,rsh,lsh){ 74              t=d[i]; 75              if(d[i-1]>t){ 76                  d[i-1]=t; 77                  flag=true; 78              } 79          } 80      } 81      return true; 82 } 83 int main() 84 { 85     std::ios::sync_with_stdio(false);//cout<<setiosflags(ios::fixed)<<setprecision(1)<<y; 86     #ifdef LOCAL 87     freopen("data.in","r",stdin); 88     freopen("data.out","w",stdout); 89     #endif 90     cin>>n;M(head,-1); 91     F(i,1,n){ 92         int a,b,c; 93         cin>>a>>b>>c; 94         addedge(b,a-1,-c); 95         if(lsh>a) lsh=a; 96         if(rsh<b) rsh=b; 97     } 98     ford(); 99     cout<<d[rsh]-d[lsh-1]<<endl;100     return 0;101 }102 /*103 5104 3 7 3105 8 10 3106 6 8 1107 1 3 1108 10 11 1109 */

poj 1201