差分约束系统 ZQUOJ 23147&&POJ 1201 Intervals

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

题意：构造一个集合，这个集合内的数字满足所给的n个条件，每个条件都是指在[a,b]内至少有c个数在集合内。问这个集合最少包含多少个点。即求至少有多少个元素在区间[a,b]内。
分析：用ti表示在[0,i-1]的范围内，要选多少个数，对于题目中所说的每个条件[a,b]内至少有c个数在集合可以表示为t(b+1)-t(a)>=c,差分约束系统的雏形已经形成。但是，题目还有一个隐藏的条件，ti表示的是在[0,i-1]的范围内，要选多少个数，数列ti是递增的，但是增量最大是1，也就是说0<=t(i)-t(i-1)<=1，这个式子等价于t(i)-t(i-1)<=1和t(i-1)-t(i)<=0。用spfa算法得到一个最长路，第一个到最后一个节点的最长路即是需要求的值。
AC代码：

View Code

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<queue>

 4 #define INF 1000000

 5 using namespace std;

 6 typedef struct

 7 {

 8     int v,w,next;

 9 }Node;

10 Node e[250000];

11 int first[50010],vis[50010],dist[50010];

12 int n,g,minx,maxx;

13 void add(int u,int v,int w)

14 {

15     e[g].v=v;

16     e[g].w=w;

17     e[g].next=first[u];

18     first[u]=g++;

19 }

20 void SPFA()

21 {

22     int i,j,t;

23     queue<int> q;

24     for(i=minx;i<=maxx;i++)

25     {

26         vis[i]=0;

27         dist[i]=-INF;

28     }

29     while(!q.empty())

30         q.pop();

31     dist[minx]=0;

32     vis[minx]=1;

33     q.push(minx);

34     while(!q.empty())

35     {

36         t=q.front();

37         q.pop();

38         vis[t]=0;

39         for(i=first[t];i!=-1;i=e[i].next)

40         {

41             j=e[i].v;

42             if(dist[j]<dist[t]+e[i].w)

43             {

44                 dist[j]=dist[t]+e[i].w;

45                 if(!vis[j])

46                 {

47                     vis[j]=1;

48                     q.push(j);

49                 }

50             }

51         }

52     }

53 }

54 int main()

55 {

56     int ai,bi,ci,i;

57     scanf("%d",&n);

58     memset(first,-1,sizeof(first));

59     g=0; minx=INF; maxx=0;

60     for(i=1;i<=n;i++)

61     {

62         scanf("%d%d%d",&ai,&bi,&ci);

63         if(ai<minx)

64             minx=ai;

65         if(bi+1>maxx)

66             maxx=bi+1;

67         add(ai,bi+1,ci);

68     }

69     for(i=minx;i<=maxx;i++)

70     {

71         add(i,i-1,-1);

72         add(i-1,i,0);

73     }

74     SPFA();

75     printf("%d\n",dist[maxx]);

76     return 0;

77 }