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
思路
发现对于整个区间的约束是可以转化成前缀和的
然后就把([a_i,b_i])这个区间的约数转化成(a_i-1和b_i)两个点，连边
然后从后向前相邻点连上权值是-1的边
从前向后连权值是0的边
然后跑一遍最长路就可以满足所有的约数条件了
注意这里的最长路，其实就是满足所有约束的最小代价
//Author: dream_maker
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
//----------------------------------------------
//typename
typedef long long ll;
//convenient for
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
//inf of different typename
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
//fast read and write
template <typename T>
void Read(T &x) {
bool w = 1;x = 0;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-') w = 0, c = getchar();
while (isdigit(c)) {
x = (x<<1) + (x<<3) + c -'0';
c = getchar();
}
if (!w) x = -x;
}
template <typename T>
void Write(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) Write(x / 10);
putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = 1e5 + 10;
struct Edge {
int v, w, nxt;
Edge(int v = 0, int w = 0, int nxt = 0):v(v), w(w), nxt(nxt) {}
} E[N * 3];
int head[N], tot = 0;
int n, a[N], b[N], c[N], maxl = 0;
int dis[N], inq[N];
void add(int u, int v, int w) {
E[++tot] = Edge(v, w, head[u]);
head[u] = tot;
}
void spfa() {
queue<int> q;
q.push(0);
fu(i, 1, maxl) dis[i] = -INF_of_int;
while (q.size()) {
int u = q.front(); q.pop();
inq[u] = 0;
for (int i = head[u]; i != -1; i = E[i].nxt) {
int v = E[i].v;
if (dis[v] < dis[u] + E[i].w) {
dis[v] = dis[u] + E[i].w;
if (!inq[v]) {
inq[v] = 1;
q.push(v);
}
}
}
}
}
int main() {
Read(n);
memset(head, -1, sizeof(head));
fu(i, 1, n) {
Read(a[i]); ++a[i];
Read(b[i]); ++b[i];
Read(c[i]);
maxl = max(maxl, max(a[i], b[i]));
add(a[i] - 1, b[i], c[i]);
}
fu(i, 1, maxl) add(i - 1, i, 0);
fu(i, 1, maxl) add(i, i - 1, -1);
spfa();
Write(dis[maxl]);
return 0;
}