POJ1201 Intervals(差分约束系统模板)

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

Source

​​Southwestern Europe 2002​​

题意：

给出n个闭区间[ai, bi]，每个区间还有个正整数ci，表示需要在区间i中至少取到ci个数。求要满足所有的n个约束条件，最少要取多少个数字。

分析：差分约束系统的入门题

书上的例题；

简单阐述一下

设S(k)为从区间[0,k]中取到的数字的个数，则

题目条件：S(bi) - S(ai - 1) >= ci。

隐含条件：0 <= S(i) - S(i - 1) <= 1，（i = 0，1，...，n-1）

（原因0~i之间选出的数看到比0~i-1之间多，每一个数只能被选一次）

为了方便计算，将所有下标向右移1位，设mx = max(bi)，则结果应为 min(S(mx) - S(0))，S(0) = 0。

要求最小值，用spfa求最长路即可。
 

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 500006, M = 5000006;
int n, m, d[N];
int Head[N], ver[M], Leng[M], Next[M], tot;
bool v[N];
 
inline void add(int x, int y, int z) {
  ver[++tot] = y;
  Leng[tot] = z;
  Next[tot] = Head[x];
  Head[x] = tot;
}
 
void spfa() {
  
  queue<int> q;
  q.push(0);
  v[0] = 1;
  d[0] = 0;
  while (q.size()) {
    int x = q.front();
    q.pop();
    v[x] = 0;
    for (int i = Head[x]; i; i = Next[i]) {
      int y = ver[i];
      if (d[y] < d[x] + Leng[i]) { ///最长路
        d[y] = d[x] + Leng[i];
        if (!v[y]) {
          v[y] = 1;
          q.push(y);
        }
      }
    }
  }
}
 
int main() {
  while(scanf("%d",&n)!=-1)
  {
  memset(v,0,sizeof(v));
  memset(d, 0xcf, sizeof(d));
  memset(Head, 0, sizeof(Head));
  tot = m = 0;
  
  for (int i = 1; i <= n; i++) {
    int x, y, z;
    scanf("%d %d %d", &x, &y, &z);
    add(x, y + 1, z)    ;
    m = max(m, y + 1);
  }
  for (int i = 1; i <= m; i++) {
    add(i - 1, i, 0);
    add(i, i - 1, -1);
  }
  spfa();
  printf("%d\n",d[m]);
  }
  return 0;
}