bzoj 1201 Intervals （考试原题 · 差分约束）

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

题解

今天考试的原题。。。然而蒟蒻我真的是太弱了不会差分。。。大佬们都是瞬秒此题而我只有暴力。。。暴力。。。
但是！现在有了大佬的亲力传授，我！也！会！了！！！
差分约束就是把一堆不等式转化成最短/最长路径，并且要注意挖掘题目的信息。如果不等式是a[i] - a[j] >= w，那么就建一条j到i权值为w的边，然后跑dis[v] < dis[u] + w即可！注意初始值赋值为正无穷或负无穷~~（一定要牢记啊。。曾经的我其实搞过这个，现在全忘了QAQ）

代码

#include <queue>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n, m, l, r, c, a[50010], vis[50010], dis[50010];struct Edge {    int v, w, next;} e[200010];int num = 0, h[200010];void add(int u, int v, int w) {    num ++;    e[num].v = v;    e[num].w = w;    e[num].next = h[u];    h[u] = num;}deque<int> q;void spfa(int s) {    while(!q.empty()) q.pop_back();    q.push_front(s), vis[1] = true, dis[s] = 0;    while(!q.empty()) {        int u=q.front(); q.pop_front(); vis[u] = false;        for(int i = h[u]; i; i =e[i].next) {            int v = e[i].v;            if(dis[v] < dis[u] + e[i].w) {                dis[v] = dis[u] + e[i].w;                if(! vis[v]) {                    if(!q.empty()&&dis[v]>dis[q.front()]) q.push_front(v);                    else q.push_back(v);                    vis[v] = true;                }            }        }    }}int main() {    n = 0;    scanf("%d", &m);    for(int i = 1; i <= m; i ++) {        int l, r, c;        scanf("%d %d %d", &l, &r, &c);        n = max(n, r); add(l - 1, r, c);    }    for(int i = 1; i <= n; i ++)        add(i - 1, i, 0), add(i, i - 1, -1);    memset(dis, -127, sizeof(dis));    spfa(0);    printf("%d", dis[n]);    return 0;}