例2.种树

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 30000+5;
const int dx[] = {-1,1,0,0,-1,-1,1,1};
const int dy[] = {0,0,-1,1,-1,1,-1,1};
using namespace std;
struct Node{
    int st,ed;
    int num;
    bool operator < (const Node &rhs)const{
        return ed<rhs.ed;
    }
}a[N];
bool vis[N];
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
        scanf("%d%d%d",&a[i].st,&a[i].ed,&a[i].num);
    sort(a+1,a+1+m);
 
    int res=0;
    for(int i=1;i<=m;i++){
        int used=0;
        for(int j=a[i].st;j<=a[i].ed;j++)//从前向后统计已有多少树
            if(vis[j])
                used++;
 
        if(used<a[i].num){//若树的个数不到当前区间所需个数
            for(int j=a[i].ed;j>=a[i].st;j--){//从后向前在当前区间种树
                if(!vis[j]){
                    vis[j]=true;
                    used++;
                    res++;
                    if(used==a[i].num)
                        break;
                }
            }
        }
    }
 
    printf("%d\n",res);
 
    return 0;
}