POJ 1201 Intervals

Intervals

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on  PKU. Original ID: 1201
64-bit integer IO format: %lld      Java class name: Main

 

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

 

解题：差分约束系统。参见 书山有路，学海无涯

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 50010;

18 struct arc{

19     int to,w,next;

20     arc(int x = 0,int y = 0,int z = 0){

21         to = x;

22         w = y;

23         next = z;

24     }

25 };

26 int n,tot,head[maxn],theMin,theMax,d[maxn];

27 arc e[maxn<<2];

28 int q[maxn<<4],he,tail;

29 bool in[maxn];

30 void add(int u,int v,int w){

31     e[tot] = arc(v,w,head[u]);

32     head[u] = tot++;

33 }

34 int spfa(){

35     for(int i = theMin; i <= theMax; i++){

36         in[i] = false;

37         d[i] = -INF;

38     }

39     d[theMin] = 0;

40     he = tail = 0;

41     q[tail++] = theMin;

42     while(he < tail){

43         int u = q[he++];

44         in[u] = false;

45         for(int i = head[u]; ~i; i = e[i].next){

46             if(d[e[i].to] < d[u]+e[i].w){

47                 d[e[i].to] = d[u]+e[i].w;

48                 if(!in[e[i].to]){

49                     in[e[i].to] = true;

50                     q[tail++] = e[i].to;

51                 }

52             }

53         }

54     }

55     return d[theMax];

56 }

57 int main() {

58     int u,v,w;

59     while(~scanf("%d",&n)){

60         memset(head,-1,sizeof(head));

61         theMin = INF;

62         theMax = -INF;

63         for(int i = tot = 0; i < n; i++){

64             scanf("%d %d %d",&u,&v,&w);

65             add(u,v+1,w);

66             theMin = min(theMin,u);

67             theMax = max(theMax,v+1);

68         }

69         for(int i = theMin; i < theMax; i++){

70             add(i,i+1,0);

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

72         }

73         printf("%d\n",spfa());

74     }

75     return 0;

76 }

View Code