Hdu 1384（差分约束）

题目链接

Intervals

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2931    Accepted Submission(s): 1067

Problem 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 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

令１..i共选ｄ[i]个数，那么有0 <= d[i + 1] - d[i] <= 1; 

并且对于每个约束a, b, c，　都有 d[b] - d[a - 1] >= c

这样就是一个线性规划问题，可以用最短路求解。

对于求最大值，就将不等式转化为求最短路中的三角不等式，即：d[u] + w >= d[v]，　然后求最短路即可。

对于求最小值，就将不等式转化为求最长路中的三角不等式，即：d[u] + w <= d[v]，　然后求最长路即可。

当然求最小值时，也可以按求最大值的方法加边，只不过这时候加的边都是反向边，从终点到起点跑最短路，然后结果取反就可以了。

Accepted Code:

 1 /*************************************************************************  2  > File Name: 1384.cpp  3  > Author: Stomach_ache  4  > Mail: [email protected]  5  > Created Time: 2014年08月26日 星期二 08时59分19秒  6  > Propose:  7  ************************************************************************/

 8 #include <queue>

 9 #include <cmath>

10 #include <string>

11 #include <cstdio>

12 #include <vector>

13 #include <fstream>

14 #include <cstring>

15 #include <iostream>

16 #include <algorithm>

17 using namespace std; 18 /*Let's fight!!!*/

19 

20 const int INF = 0x3f3f3f3f; 21 const int MAX_N = 50050; 22 typedef pair<int, int> pii; 23 vector<pii> G[MAX_N]; 24 int n, d[MAX_N]; 25 bool inq[MAX_N]; 26 

27 void AddEdge(int u, int v, int w) { 28  G[u].push_back(pii(v, w)); 29 } 30 

31 void spfa(int s) { 32       queue<int> Q; 33     memset(d, 0x3f, sizeof(d)); 34     memset(inq, false, sizeof(inq)); 35     d[s] = 0; 36     inq[s] = true; 37  Q.push(s); 38     while (!Q.empty()) { 39           int u = Q.front(); Q.pop(); inq[u] = false; 40         for (int i = 0; i < G[u].size(); i++) { 41               int v = G[u][i].first, w = G[u][i].second; 42             if (d[u] + w < d[v]) { 43                   d[v] = d[u] + w; 44                 if (!inq[v]) Q.push(v), inq[v] = true; 45  } 46  } 47  } 48 } 49 

50 int main(void) { 51       while (~scanf("%d", &n)) { 52           for (int i = 0; i <= 50005; i++) G[i].clear(); 53         int s = INF, t = -1; 54         for (int i = 0; i < n; i++) { 55               int a, b, c; 56             scanf("%d %d %d", &a, &b, &c); 57             b++; 58             s = min(s, a); t = max(t, b); 59             AddEdge(b, a, -c); 60  } 61         for (int i = s; i < t; i++) AddEdge(i, i + 1, 1), AddEdge(i + 1, i, 0); 62 

63  spfa(t); 64 

65         printf("%d\n", -d[s]); 66  } 67 

68     return 0; 69 }