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:

 /*************************************************************************
     > File Name: 1384.cpp
     > Author: Stomach_ache
     > Mail: [e-mail]
     > Created Time: 2014年08月26日 星期二 08时59分19秒
     > Propose:
  ************************************************************************/
 #include <queue>
 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <vector>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 /*Let's fight!!!*/

 const int INF = 0x3f3f3f3f;
 const int MAX_N = ;
 typedef pair<int, int> pii;
 vector<pii> G[MAX_N];
 int n, d[MAX_N];
 bool inq[MAX_N];

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

 void spfa(int s) {
       queue<int> Q;
     memset(d, 0x3f, sizeof(d));
     memset(inq, false, sizeof(inq));
     d[s] = ;
     inq[s] = true;
     Q.push(s);
     while (!Q.empty()) {
           int u = Q.front(); Q.pop(); inq[u] = false;
         for (int i = ; i < G[u].size(); i++) {
               int v = G[u][i].first, w = G[u][i].second;
             if (d[u] + w < d[v]) {
                   d[v] = d[u] + w;
                 if (!inq[v]) Q.push(v), inq[v] = true;
             }
         }
     }
 }

 int main(void) {
       while (~scanf("%d", &n)) {
           for (int i = ; i <= ; i++) G[i].clear();
         int s = INF, t = -;
         for (int i = ; i < n; i++) {
               int a, b, c;
             scanf("%d %d %d", &a, &b, &c);
             b++;
             s = min(s, a); t = max(t, b);
             AddEdge(b, a, -c);
         }
         for (int i = s; i < t; i++) AddEdge(i, i + , ), AddEdge(i + , i, );

         spfa(t);

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

     return ;
 }