POJ1201 Intervals差分约束系统（最短路）

时间：2021-08-21 04:57:11

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

题意是给你n个闭区间，形如[ai,bi]，在这n个区间的每一个整点位置放东西，每个区间至少要放ci个，求总的最少要放的个数
了解过查分约束系统之后，比较容易知道是差分系统的最小差问题，那么就是找全形如 y-x>=k的不等式，然后再求一个最长路
关键是建图问题：p[i]表示1～i放的东西个数，那么对于每一个区间，就要满足p[bi] - p[ai-1]>=ci，即拉一条(ai-1)->bi的有向边
但是有一个隐藏的约束条件，由于是整点放，那么p[x] - p[x-1] <= 1 && p[x] - p[x-1] >= 0  即p[x-1]-p[x]>=-1 && p[x]-p[x-1]>=0
另外，由于区间可以从0开始，那么把所有的区间整体右移2个位置，就是从1开始的了
#include <cstdio>

#include <cstring>

#include <algorithm>

#include <queue>

#include <stack>

#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f;

const int N = 5e4 + ;

struct Edge {

    int v, cost;

    Edge() {}

    Edge(int v, int cost) : v(v), cost(cost) {}

};

vector<Edge> E[N];

void add(int u, int v, int w) {

    E[u].push_back(Edge(v, w));

}

bool vis[N];

int d[N];

void SPFA(int st, int n) {

    memset(vis, false, sizeof vis);

    for(int i = ; i <= n; ++i) d[i] = -INF;

    d[st] = ;

    vis[st] = true;

    queue<int> que;

    while(!que.empty()) que.pop();

    que.push(st);

    while(!que.empty()) {

        int u = que.front();

        que.pop();

        vis[u] = false;

        int sx = E[u].size();

        for(int i = ; i < sx; ++i) {

            int v = E[u][i].v;

            int w = E[u][i].cost;

            if(d[v] < d[u] + w) {

                d[v] = d[u] + w;

                if(!vis[v]) {

                    vis[v] = true;

                    que.push(v);

                }

            }

        }

    }

}

int main() {

    int n;

    while(~scanf("%d", &n)) {

        int a, b, c, m = , st = INF;

        for(int i = ; i < n; ++i) {

            scanf("%d%d%d", &a, &b, &c);

            a+=; b+=;

            add(a - , b, c);

            m = max(m, b);

            st = min(st, a - );

        }

        for(int i = ; i <= m; ++i) {

            add(i, i - , -);

            add(i - , i, );

        }

        SPFA(st, m);

        printf("%d\n", d[m]);

    }

    return ;

}

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