poj1201/zoj1508/hdu1384 Intervals(差分约束)

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Intervals

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.

Write a program that:

> reads the number of intervals, their endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi - ai + 1.

Process to the end of file.

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：

对于一个序列，有n个描述，[ai,bi,ci]分别表示在区间[ai,bi]上，至少有ci个数属于该序列。输出满足这n个条件的最短的序列（即包含的数字个数最少）

分析：

设s[i]表示从0到i中有s[i]个数属于这个序列。

那么，对于每个描述，都可以得到一个方程s[bi]-s[ai-1]>=ci

同时由于每个数至多取一次，那么有s[i]-s[i-1]<=1,即s[i-1]-s[i]>=-1;

另外还有s[i]-s[i-1]>=0;

由这三个方程组建图，利用最长路即可求出。

 1 #include <iostream>

 2 #include <cstring>

 3 #include <cstdio>

 4 #include <cstdlib>

 5 #include <algorithm>

 6 #include <queue>

 7 using namespace std;

 8 typedef long long ll;

 9 typedef pair<int,int> PII;

10 #define MAXN 50010

11 #define REP(A,X) for(int A=0;A<X;A++)

12 #define INF  0x7fffffff

13 #define CLR(A,X) memset(A,X,sizeof(A))

14 struct node {

15     int v,d,next;

16 }edge[3*MAXN];

17 int head[MAXN];

18 int e=0;

19 void init()

20 {

21     REP(i,MAXN)head[i]=-1;

22 }

23 void add_edge(int u,int v,int d)

24 {

25     edge[e].v=v;

26     edge[e].d=d;

27     edge[e].next=head[u];

28     head[u]=e;

29     e++;

30 }

31 bool vis[MAXN];

32 int dis[MAXN];

33 void spfa(int s){

34     CLR(vis,0);

35     REP(i,MAXN)dis[i]=i==s?0:INF;

36     queue<int>q;

37     q.push(s);

38     vis[s]=1;

39     while(!q.empty())

40     {

41         int x=q.front();

42         q.pop();

43         int t=head[x];

44         while(t!=-1)

45         {

46             int y=edge[t].v;

47             int d=edge[t].d;

48             t=edge[t].next;

49             if(dis[y]>dis[x]+d)

50             {

51                 dis[y]=dis[x]+d;

52                 if(vis[y])continue;

53                 vis[y]=1;

54                 q.push(y);

55             }

56         }

57         vis[x]=0;

58     }

59 }

60 int main()

61 {

62     ios::sync_with_stdio(false);

63     int n;

64     int u,v,d;

65     int ans=0;

66     int maxn=0;

67     int minn=MAXN;

68     while(scanf("%d",&n)!= EOF&&n)

69     {

70         e=0;

71         init();

72         REP(i,n)

73         {

74             scanf("%d%d%d",&u,&v,&d);

75             add_edge(u,v+1,-d);

76             maxn=max(maxn,v+1);

77             minn=min(u,minn);

78         }

79         for(int i=minn;i<maxn;i++){

80             add_edge(i+1,i,1);

81             add_edge(i,i+1,0);

82         }

83         spfa(minn);

84         cout<<-dis[maxn]<<endl;

85     }

86     return 0;

87 }

代码君