POJ 1201 Intervals 【差分约束】

最新推荐文章于 2021-05-04 00:09:04 发布

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本文介绍了一个关于区间覆盖的问题，并提供了一种通过编程解决该问题的方法。具体来说，需要找到一个整数集合，使得该集合与给定的多个闭合区间至少有指定数量的交集元素，目标是最小化这个集合的大小。

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Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 28690		Accepted: 11068

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

Sample Output

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int MAX = 5e5 + 7;
const int INF = 0x3f3f3f3f;
struct node
{
    int to, c;
};
int n, mx, mn;
vector <node> V[MAX];
queue <int> q;
void add(int x, int y, int z)
{
    V[x].push_back({y, z});
}
int dis[MAX], vis[MAX];
void spfa()
{
    fill(dis + mn + 1, dis + mx + 1, -INF);
    vis[mn] = 1;
    q.push(mn);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        int len = V[u].size();
        for(int i = 0; i < len; i++)
        {
            int to = V[u][i].to;
            int c = V[u][i].c;
            if(dis[to] < dis[u] +c)
            {
                dis[to] = dis[u] + c;
                if(!vis[to])
                {
                    q.push(to);
                    vis[to] = 1;
                }
            }
        }
    }
    printf("%d\n", dis[mx]);
}
int main()
{
    scanf("%d", &n);
    for(int i = 0, x, y, z; i < n; i++)
    {
        scanf("%d%d%d", &x, &y, &z);
        add(x, y + 1, z);
        mx = max(y, mx);
        mn = min(x, mn);
    }
    mx++;
    for(int i = mn; i < mx; i++)
    {
        add(i, i + 1, 0);
        add(i + 1, i, -1);
    }
    spfa();
    return 0;
}