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

53 7 38 10 36 8 11 3 110 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 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             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
             
               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
              
                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;}