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: sudaweitong@gmail.com
    > 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 = 50050;
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] = 0;
    inq[s] = true;
    Q.push(s);
    while (!Q.empty()) {
          int u = Q.front(); Q.pop(); inq[u] = false;
        for (int i = 0; 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 = 0; i <= 50005; i++) G[i].clear();
        int s = INF, t = -1;
        for (int i = 0; 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 + 1, 1), AddEdge(i + 1, i, 0);

        spfa(t);

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

    return 0;
}