Codeforces Round #436 (Div. 2) E. Fire

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Description

Polycarp is in really serious trouble — his house is on fire! It’s time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di — the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti ≥ di, then i-th item cannot be saved.

Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b — in ta + tb seconds after fire started.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of items in Polycarp’s house.

Each of the following n lines contains three integers ti, di, pi (1 ≤ ti ≤ 20, 1 ≤ di ≤ 2 000, 1 ≤ pi ≤ 20) — the time needed to save the item i, the time after which the item i will burn completely and the value of item i.

Output

In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m — the number of items in the desired set. In the third line print m distinct integers — numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.

Examples

input
3
3 7 4
2 6 5
3 7 6
output
11
2
2 3 
input
2
5 6 1
3 3 5
output
1
1
1

题目大意

有n件物品，拯救第i件需要时间ti，价值为pi，到达时间di时物品会被烧毁。问能拯救的最大价值及拯救的物品标号。

解题思路

先将所有物品按照烧毁的时间排序，然后01背包，记录路径。
状态转移方程：
j>=a[i].d||a[i].t>j时，dp[i][j]=dp[i-1][j]；
其他情况下， dp[i][j]=max(dp[i-1][j],dp[i-1][j-a[i].t]+a[i].p)。

代码实现

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
#define maxn 2007
vector<int>res;
struct node
{int t;int d;int p;int index;
} a[107];
int dp[107][maxn];
int mark[107][maxn];
bool cmp(node x ,node y)
{return x.d<y.d;
}
int main()
{int n,ans,pos;while(~scanf("%d",&n)){ans=0,pos=0;memset(dp,0,sizeof(dp));memset(mark,-1,sizeof(mark));for(int i=1; i<=n; i++){scanf("%d %d %d",&a[i].t,&a[i].d,&a[i].p);a[i].index=i;}sort(a+1,a+n+1,cmp);for(int i=1; i<=n; i++){for(int j=1; j<=2000;j++){dp[i][j]=dp[i-1][j];if(!(j>=a[i].d||a[i].t>j)){if(dp[i-1][j-a[i].t]+a[i].p>dp[i][j]){dp[i][j]=dp[i-1][j-a[i].t]+a[i].p;mark[i][j]=i;}}}}for(int j=1;j<=2000;j++){if(ans<dp[n][j]){ans=dp[n][j];pos=j;}}printf("%d\n",ans);res.clear();int maxx=n;while(pos>0&&maxx>0){if(mark[maxx][pos]>0){res.push_back(a[mark[maxx][pos]].index);pos-=a[mark[maxx][pos]].t;maxx--;}elsemaxx--;}int length=res.size();reverse(res.begin(),res.end());printf("%d\n",length);for(int i=0; i<length; i++)printf(i==length-1?"%d\n":"%d ",res[i]);}return 0;
}