CF864E：E. Fire（01背包 & 路径输出）

最新推荐文章于 2023-06-03 07:30:00 发布

原创最新推荐文章于 2023-06-03 07:30:00 发布 · 496 阅读

2 ·

CC 4.0 BY-SA版权

背包问题专栏收录该内容

30 篇文章

订阅专栏

本文介绍了一种在火灾场景下，如何最大化救援物品价值的算法。通过动态规划的方法，在考虑物品救援时间和燃烧时限的情况下，确定最优救援顺序。

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.

题意：火灾时有N件物品可以救出，每一件有三个值a，b，c，表示需要救它要a时间，且在b时间该物品就立刻烧毁，c为它的价值，问最多可以救出多少价值，并输出救的顺序。

思路：明显的背包dp，由于本题有“b”属性的限制，b小的物品不能从b大的状态转移过来，于是需要对b排序，先处理b小的物品。路径输出的话范围小可以用vector直接搞，也可以用数组标记一下。

①

# include <iostream>
# include <cstdio>
# include <cstring>
# include <vector>
# include <algorithm>
using namespace std;
const int maxn = 2003;
int a[maxn], b[maxn], c[maxn], dp[maxn];
int id[maxn];
vector<int>v[maxn];
bool cmp(int x, int y)
{
    return b[x] < b[y];
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; ++i)
        scanf("%d%d%d",&a[i],&b[i],&c[i]),id[i] = i;
    sort(id, id+n, cmp);
    for(int k=0; k<n; ++k)
    {
        int i = id[k];
        if(a[i]>=b[i]) continue;
        for(int j=b[i]-1; j>=a[i]; --j)
        {
            if(dp[j-a[i]]+c[i] > dp[j])
            {
                dp[j] = dp[j-a[i]]+c[i];
                v[j].clear();
                v[j] = v[j-a[i]];
                v[j].emplace_back(i);
            }
        }
    }
    int imax=0, num;
    for(int i=0; i<=2000;++i)
        if(dp[i] > imax)
            imax = dp[i], num=i;
    printf("%d\n",imax);
    printf("%d\n",v[num].size());
    for(auto i : v[num])
        printf("%d ",i+1);
    return 0;
}

②

# include <iostream>
# include <cstdio>
# include <cstring>
# include <stack>
# include <algorithm>
using namespace std;
const int maxn = 2003;
int a[maxn], b[maxn], c[maxn], dp[maxn];
int id[maxn], cho[maxn], way[103][maxn];
bool cmp(int x, int y)
{
    return b[x] < b[y];
}
int main()
{
    int n, imax, ans=0, num;
    scanf("%d",&n);
    for(int i=1; i<=n; ++i)
        scanf("%d%d%d",&a[i],&b[i],&c[i]),id[i] = i;
    sort(id+1, id+n+1, cmp);
    for(int k=1; k<=n; ++k)
    {
        int i = id[k];
        if(a[i]>=b[i]) continue;
        for(int j=b[i]-1; j>=a[i]; --j)
        {
            if(dp[j-a[i]]+c[i] > dp[j])
            {
                dp[j] = dp[j-a[i]]+c[i];
                cho[j] = i;
                way[i][j] = cho[j-a[i]];
            }
            if(dp[j] > ans)
            {
                ans = dp[j];
                imax = j;
                num = i;
            }
        }
    }
    printf("%d\n",ans);
    if(ans == 0) return 0*puts("0");
    stack<int>st;
    while(imax && num)
    {
        st.push(num);
        int tmp = imax;
        imax -= a[num];
        num = way[num][tmp];
    }
    printf("%d\n",st.size());
    while(!st.empty())
    {
        printf("%d ",st.top());
        st.pop();
    }
    return 0;
}