Russian Dolls

Time Limit: 2000ms, Special Time Limit:5000ms, Memory Limit:65536KB

Total submit users: 12, Accepted users: 9

Problem 12895 : No special judgement

Problem description

Maybe you know the famous russian souvenir Russian Dolls. It looks like a set of nested wooden dolls. A doll with a smaller size is placed inside a bigger one. Let's consider all dolls taken apart. Each doll has an outer volume outi which is the volume it occupies in space and an inner volume ini - the volume of the empty space inside the doll. You may assume that you can put one doll inside another if the outer volume of the first doll is strictly less than the inner volume of the second one. If two or more dolls are inside another one they can't lie one near the other, they must be nested.

For each doll the cost of unit of empty space - costi is known. You must pay exactly costi for each unit of empty space which directly belongs to the i-th doll (but not to ones inside it). You may arrange the dolls the way you want as long as you are not contradicting the rules. The objective is to find an arrangement of nesting the dolls (not necessarily all of them) such that the overall cost you have to pay is minimized.

Input

First line contains an integer N (1 ≤ N ≤ 1000) which is the number of dolls you have. The i-th of the next N lines contains three integers outi, ini, costi (1 ≤ ini < outi ≤ 1000, 1 ≤ costi ≤ 1000), which are the outer volume, inner volume and the empty space cost of the i-th doll.

Output

Single integer P which is the minimum possible cost you should pay.

Sample Input

3
5 4 1
4 2 2
3 2 1

Sample Output

7

#include<iostream>
#include<cstring>//一個是思路問題，一個是要是升序問題，一個是排序後它會變序，要新建一個數組排序 
#include<cstdio>
#include<algorithm>//求最大值時別忘了改值 
#define inf 999999
#define maxn 1010
using namespace std;
int outi[maxn];
int ini[maxn];
struct node
{
	int out,in,cos;
}str[maxn];
bool v[maxn]; 
int temp[maxn];
bool cmpa(int a,int b)
{
	return str[a].out>str[b].out;
}
bool cmpb(int a,int b)//這是升序吧 
{
	if(str[a].in==str[b].in)
	return str[a].cos>str[b].cos;
	return str[a].in>str[b].in;
}
int main()//break，初始化 
{
	int T;
	while(scanf("%d",&T)!=EOF)
	{
		for(int i=1;i<=T;i++)
		{
			cin>>str[i].out>>str[i].in>>str[i].cos;
			ini[i]=i;outi[i]=i;
		}
		sort(outi+1,outi+T+1,cmpa);//排序問題，升序 
		sort(ini+1,ini+T+1,cmpb);
		int ans=0;
		memset(v,0,sizeof(v));
		for(int i=1;i<=T;i++)
		{
			int x=0;
			for(int j=1;j<=T;j++)
			{
				if(str[outi[i]].out<str[ini[j]].in&&!v[ini[j]])
			 
				temp[++x]=ini[j];
			}
			//cout<<x<<endl;
			if(x>=1)
			{
			
			int vmax=0;
			int res;
			for(int k=1;k<=x;k++)
			{
			
				int tep=str[temp[k]].cos*str[outi[i]].out;//減的量 
				//cout<<tep<<endl;
				if(vmax<tep)
				{
					vmax=tep;
					res=temp[k];//這裏一開始忘了改值 
				}
			}
			v[res]=1;
			//cout<<res<<endl;
			ans+=str[res].cos*str[outi[i]].out;
		    }
		}
		int result=0;
		for(int i=1;i<=T;i++)
		{
			result+=str[i].cos*str[i].in;
		}
		
		result-=ans;
		//cout<<ans<<endl;
		cout<<result<<endl;
	}
}