Captain Fint is involved in another treasure hunt, but have found only one strange problem. The problem may be connected to the treasure’s location or may not. That’s why captain Flint decided to leave the solving the problem to his crew and offered an absurdly high reward: one day off. The problem itself sounds like this…

There are two arrays 𝑎 and 𝑏 of length 𝑛. Initially, an 𝑎𝑛𝑠 is equal to 0 and the following operation is defined:

Choose position 𝑖 (1≤𝑖≤𝑛);
Add 𝑎𝑖 to 𝑎𝑛𝑠;
If 𝑏𝑖≠?1 then add 𝑎𝑖 to 𝑎𝑏𝑖.
What is the maximum 𝑎𝑛𝑠 you can get by performing the operation on each 𝑖 (1≤𝑖≤𝑛) exactly once?

Uncle Bogdan is eager to get the reward, so he is asking your help to find the optimal order of positions to perform the operation on them.

Input
The first line contains the integer 𝑛 (1≤𝑛≤2?105) — the length of arrays 𝑎 and 𝑏.

The second line contains 𝑛 integers 𝑎1,𝑎2,…,𝑎𝑛 (?106≤𝑎𝑖≤106).

The third line contains 𝑛 integers 𝑏1,𝑏2,…,𝑏𝑛 (1≤𝑏𝑖≤𝑛 or 𝑏𝑖=?1).

Additional constraint: it’s guaranteed that for any 𝑖 (1≤𝑖≤𝑛) the sequence 𝑏𝑖,𝑏𝑏𝑖,𝑏𝑏𝑏𝑖,… is not cyclic, in other words it will always end with ?1.

Output
In the first line, print the maximum 𝑎𝑛𝑠 you can get.

In the second line, print the order of operations: 𝑛 different integers 𝑝1,𝑝2,…,𝑝𝑛 (1≤𝑝𝑖≤𝑛). The 𝑝𝑖 is the position which should be chosen at the 𝑖-th step. If there are multiple orders, print any of them.

Examples
inputCopy
3
1 2 3
2 3 -1
outputCopy
10
1 2 3
inputCopy
2
-1 100
2 -1
outputCopy
99
2 1
inputCopy
10
-10 -1 2 2 5 -2 -3 -4 2 -6
-1 -1 2 2 -1 5 5 7 7 9
outputCopy
-9
3 5 6 1 9 4 10 7 8 2

題意：
每次操作你可以選擇$i$，并且獲得$a[i]$的值。同時如果$b[i]!=?1$，$a[b[i]]$加上$a[i]$。每個$i$只能選擇一次，問最大能獲得多大的值。
保證$i,b[i],b[b[i]]...$不是環。

思路：
因為$i,b[i],b[b[i]]...$不是環，所以可以$i$向$b[i]$連邊，則構成了一棵樹。對于所有$b[i]=?1,i$為該樹根節點。

因為每次選擇完$i$會對$b[i]$有影響，所以我們肯定希望影響是正面的。也就是如果$a[i]$是正數，毫無疑問得算上這個影響。否則就把$i$放到后面選擇。

這樣貪心算法就出來了，遞歸的考慮，考慮所有子樹的選擇，如果$a[v]$大于0，那么就早于$x$選擇，否則后于$x$選擇。并且$a[x]$加上所有大于0的$a[v]$。

路徑等同于dfs的順序，當子節點的$a[v]$為負數時，則先輸出$x$，再輸出子樹v的節點。

#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <vector> #include <queue> #include <iostream> #include <map> #include <string> #include <set> #include <deque> typedef long long ll; using namespace std;const int maxn = 2e5 + 7; const ll INF = 0x3f3f3f3f3f3f3f3f; ll a[maxn],b[maxn]; vector<int>G[maxn]; ll f[maxn]; deque<int>res;void DP(int x) {f[x] = a[x];ll sum = 0;for(int i = 0;i < G[x].size();i++) {int v = G[x][i];DP(v);f[x] += f[v];if(a[v] >= 0) sum += a[v];}a[x] += sum;f[x] += sum; }int cmp(int x,int y) {return a[x] > a[y]; }void dfs2(int x) {sort(G[x].begin(),G[x].end(),cmp);int flag = 0;for(int i = 0;i < G[x].size();i++) {int v = G[x][i];if(a[v] < 0 && flag == 0) {res.push_back(x);flag = 1;}dfs2(v);}if(!flag) res.push_back(x); }int main() {int n;scanf("%d",&n);for(int i = 1;i <= n;i++) {scanf("%lld",&a[i]);}for(int i = 1;i <= n;i++) {scanf("%lld",&b[i]);}ll ans = 0;for(int i = 1;i <= n;i++) {if(b[i] != -1) {G[b[i]].push_back(i);}}for(int i = 1;i <= n;i++) {if(b[i] == -1) {DP(i);ans += f[i];}}for(int i = 1;i <= n;i++) {if(b[i] == -1) {dfs2(i);}}printf("%lld\n",ans);while(!res.empty()) {printf("%d ",res.front());res.pop_front();}return 0; }

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