diff --git a/hackpack/approaches/dp/problems/picnic/Makefile b/hackpack/approaches/dp/problems/picnic/Makefile
new file mode 100644
index 0000000..2aecb3c
--- /dev/null
+++ b/hackpack/approaches/dp/problems/picnic/Makefile
@@ -0,0 +1,8 @@
+CXX=g++
+CXXFLAGS=-std=gnu++0x
+all: picnic.o
+	$(CXX) $^ -o picnic.exe
+test: all
+	./picnic.exe < picnic.in
+clean:
+	-rm picnic.exe *.o
diff --git a/hackpack/approaches/dp/problems/picnic/picnic.cpp b/hackpack/approaches/dp/problems/picnic/picnic.cpp
new file mode 100644
index 0000000..e25515a
--- /dev/null
+++ b/hackpack/approaches/dp/problems/picnic/picnic.cpp
@@ -0,0 +1,68 @@
+#include<iostream>
+#include<algorithm>
+#include<map>
+
+using namespace std; 
+
+class employee{
+    public:
+        int id;
+        int value; 
+        employee(int id,int value): id(id), value(value){};
+};
+
+//key is supervisor,
+multimap<int, employee> company;
+map<int,int> max_present;
+map<int,int> max_not_present;
+
+int solve(multimap<int, employee>::iterator r, bool present){
+    //No need to check both present and not present because they 
+    //are evaluated at the same time
+    if(max_present.find(r->first) != max_present.end()){
+        return ((present)? max_present[r->first]: max(max_present[r->first],
+                    max_not_present[r->first]));
+    }
+
+    //Compute the best solution to the subproblem.
+    int sum_present=0, sum_not_present=0;
+    int supervisor = r->second.id;
+    auto subordinates = company.equal_range(supervisor);
+
+    //Base case the employee has no subordinates
+    if(subordinates.first == company.end() && subordinates.second == company.end()){
+        return ((present)? 0 : r->second.value );
+    }
+    
+    //Otherwise compute the present vs not present sums
+    for(auto i = subordinates.first; i!= subordinates.second; i++){
+        int not_present = solve(i,true);
+        int present = solve(i,false);
+        sum_present += present;
+        sum_not_present += max(present, not_present);
+    }
+
+    //Store off the best results for present vs not present
+    max_present[r->second.id] = sum_present;
+    max_not_present[r->second.id] = sum_not_present;
+    return ((present)? sum_not_present: max(sum_not_present, sum_present));
+
+
+}
+
+int main(){
+    int id, value, supervisor;
+    while(cin >> id >> value >> supervisor){
+        company.insert(make_pair(supervisor, employee(id, value)));
+    }
+    cout << max(solve(company.find(0),true) ,solve(company.find(0),false)) << endl;
+
+
+    /*
+    for(auto i: company){
+        cout << "[" << i.first << ", " << i.second.id << ", " << i.second.value
+            << "]" << endl;
+    }
+    */
+}
+
diff --git a/hackpack/approaches/dp/problems/picnic/picnic.in b/hackpack/approaches/dp/problems/picnic/picnic.in
new file mode 100644
index 0000000..ea18db3
--- /dev/null
+++ b/hackpack/approaches/dp/problems/picnic/picnic.in
@@ -0,0 +1,3 @@
+1 2 3
+2 3 3
+3 3 1