POJ1201 Intervals查分约束系统（最短路）

Posted 2020-06-24 A.T.R_p.orc

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 + 10;
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 = 1; i <= n; ++i) d[i] = -INF;
    d[st] = 0;
    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 = 0; 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 = 0, st = INF;
        for(int i = 0; i < n; ++i) {
            scanf("%d%d%d", &a, &b, &c);
            a+=2; b+=2;
            add(a - 1, b, c);
            m = max(m, b);
            st = min(st, a - 1);
        }
        for(int i = 2; i <= m; ++i) {
            add(i, i - 1, -1);
            add(i - 1, i, 0);
        }
        SPFA(st, m);
        printf("%d\n", d[m]);
    }
    return 0;
}

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