Codeforces 864E Fire【排序+背包dp】

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面对房屋火灾，Polycarp需要制定策略，在有限时间内抢救家中物品。每件物品有其抢救时间和价值，且超过一定时间将无法抢救。任务是找出最大价值的物品集合及抢救顺序，涉及动态规划算法。

E. Fire
time limit per test 2 seconds
memory limit per test 256 megabytes
input standard input
output standard output
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 tiseconds 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
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2seconds. Thus, the total value of the saved items will be 6 + 5 = 11.

In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3seconds, but this item will already be completely burned by this time.

题目大意：

给出N个物品，每个物品拿取需要ti时间，di之间之后这个物品就不能拿了，物品的价值为pi，问怎样拿能够拿最多价值的物品，问这些物品拿取的顺序。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>

using namespace std;

struct node
{
    int t,d,p,id;
}itm[105];

bool cmp(node x,node y)
{
    return x.d<y.d;
}
int n;

int dp[125][2005];
int id[125][2005];
int pre[125][2005];

int cnt;
int anss[125];
int main()
{
    while(~scanf("%d",&n))
    {
        cnt=0;
        for(int j=0;j<=120;j++)
        for(int i=0;i<=2001;i++)
        {
            dp[j][i]=0;
            id[j][i]=0;
            pre[j][i]=0;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&itm[i].t,&itm[i].d,&itm[i].p);
            itm[i].id=i;
        }
        sort(itm+1,itm+(1+n),cmp);

        for(int i=1;i<=n;i++)
        {
            for(int j=2000;j>=0;j--)
            {
                dp[i][j]=dp[i-1][j];
                id[i][j]=0;
                pre[i][j]=j;

                if(j<itm[i].d && (j-itm[i].t)>=0)
                {
                    if(dp[i][j]<dp[i-1][j-itm[i].t]+itm[i].p)
                    {
                        dp[i][j]=dp[i-1][j-itm[i].t]+itm[i].p;
                        id[i][j]=itm[i].id;
                        pre[i][j]=j-itm[i].t;
                    }
                }
            }
        }
        int ansid=0;
        int ans=0;
        for(int i=0;i<=2000;i++)
            if(ans<dp[n][i])
            {
                ans=dp[n][i];
                ansid=i;
            }
        int tid=n;
        while(ansid!=0)
        {
            if(id[tid][ansid]!=0)
            {
               anss[cnt]=id[tid][ansid];
               cnt++;
            }
            ansid=pre[tid][ansid];
            tid--;
        }
        printf("%d\n",ans);
        printf("%d\n",cnt);
        for(int i=cnt-1;i>=0;i--)
            printf("%d ",anss[i]);
        printf("\n");
    }
}