Worksession progress

Makefile

@@ -10,6 +10,6 @@ project5:	project5.cc province.cc province.h
 # test the code against an expected output file
 # this test should grow into a spaceship and end with blinkers
 test-requirement1:	project5
-	./project5 < test-data/requirement1.in > test-data/project5.out
-	diff test-data/project5.out test-data/requirement1.out > test.diff
-	wc -l test.diff
+	./project5 < test-data/t0502-combo.in > test-data/project5.out
+	diff test-data/project5.out test-data/shortest.out > test.diff
+	rm test-data/project5.out

project5.cc

@@ -37,29 +37,11 @@ int main(int argc, char *argv[]) {
         // create a new graph for each loop, which will read all of the
         // corresponding data per graph
         Province theProvince(cin);
-
-        std::cout << std::endl;
-        std::cout << "------------------------------------------------" << std::endl;
-        std::cout << "---------------- New DataSet: ------------------" << std::endl;
-        std::cout << "------------------------------------------------" << std::endl;
-        std::cout << std::endl;
-
         theProvince.printAll(0, cout);
-
-
         theProvince.printShortestPath(std::cout);
-
-        std::cout << std::endl;
-        std::cout << "------------------------------------------------" << std::endl;
-        std::cout << "------------------------------------------------" << std::endl;
-        std::cout << std::endl;
-
         theProvince.minSpan(std::cout);
-
-        std::cout << std::endl;
-        std::cout << "------------------------------------------------" << std::endl;
-        std::cout << "------------------------------------------------" << std::endl;
         std::cout << std::endl;
-
+        std::cout << "------------------------------------------------------------------";
+        std::cout << endl;
     }
 }

province.cc

@@ -79,23 +79,20 @@ void Province::printAll(int start, std::ostream & output) {
   queue <int> toVisit; // use queue to keep track of which town to visit next
   toVisit.push(start);
   scheduled[start] = true;
-  output << "The input data is :" << endl << endl;
+  output << "The input data is:" << endl << endl;
 
   // Visit every town in the queue
   while (!toVisit.empty()) {
     int current = toVisit.front();
     toVisit.pop();
 
-    output << "   ";
     output << _towns[current]._name << endl;
 
     // Add current town's(vertex's) unscheduled neighbors to the queue
     for (Town::RoadList::iterator neighbor = _towns[current]._roads.begin();
       neighbor != _towns[current]._roads.end(); neighbor++) {
         std::string neighborName = _towns[neighbor->_head]._name;
-
-        output << "         ";
-        output << neighborName << " " << neighbor->_length << " mi";
+        output << "        " << neighborName << " " << neighbor->_length << " mi";
 
         // if the type is bridge, then add to output
         if (neighbor ->_isBridge) {
@@ -113,7 +110,7 @@ void Province::printAll(int start, std::ostream & output) {
         }
       }
   }
-  output << endl << endl;
+  output << endl;
 }
 
 int Province::smallest(double dist[], std::list <int> toVisit,
@@ -147,7 +144,7 @@ void Province::printShortestPath(std::ostream & output) const {
         return;
     }
 
-    output << "The shortest routes from " + _towns[0]._name;
+    output << "The shortest paths from " + _towns[0]._name;
     output << " are:" << std::endl << std::endl;
 
     // keeps track of the index of the predecessor to each
@@ -194,7 +191,7 @@ void Province::printShortestPath(std::ostream & output) const {
 
     // print out the data for each non capital town
     for (int i = 1; i < _numberOfTowns; i++) {
-        output << "      " << "The shortest route from " + _towns[0]._name;
+        output << "    " << "The shortest path from " + _towns[0]._name;
         output << " to " + _towns[i]._name + " is " << dist[i];
         output << " mi:" << std::endl;
 
@@ -214,11 +211,12 @@ void Province::printShortestPath(std::ostream & output) const {
 
         // print out the names for each entry in the stack
         while (!predecessors.empty()) {
-            output << "            " << _towns[predecessors.top()]._name;
+            output << "        " << _towns[predecessors.top()]._name;
             output << std::endl;
             predecessors.pop();
         }
     }
+  cout << endl;  
 }
 
 /**
@@ -308,11 +306,11 @@ void Province::minSpan(std::ostream & output) const {
     }
 
     output << "The road upgrading goal can be achieved at minimal cost by upgrading:";
-    output << std::endl << std::endl;
+    output << std::endl;
 
     // Print names of towns in minimum spanning tree of province
     for (int i = 0; i < minSpanTree.size(); i++) {
-        output << "      ";
+        output << "    ";
         output << _towns[minSpanTree[i]._head]._name;
         output << " to ";
         output << _towns[minSpanTree[i]._tail]._name << std::endl;

test-data/project5.out

@@ -1,84 +1,42 @@
+The input data is:
 
-------------------------------------------------
----------------- New DataSet: ------------------
-------------------------------------------------
+Wenham
+        Hamilton 1 mi
+Hamilton
+        Wenham 1 mi
 
-The input data is :
+The shortest paths from Wenham are:
 
-   Salem
-         Beverly 2.4 mi via bridge
-         Danvers 3.7 mi via bridge
-         Lynn 4.9 mi
-   Beverly
-         Danvers 2.9 mi
-         Salem 2.4 mi via bridge
-         Wenham 5.2 mi
-   Danvers
-         Beverly 2.9 mi
-         Wenham 4.2 mi
-         Salem 3.7 mi via bridge
-   Lynn
-         Salem 4.9 mi
-   Wenham
-         Beverly 5.2 mi
-         Danvers 4.2 mi
-
-
-The shortest routes from Salem are:
-
-      The shortest route from Salem to Wenham is 7.6 mi:
-            Salem
-            Beverly
-            Wenham
-      The shortest route from Salem to Beverly is 2.4 mi:
-            Salem
-            Beverly
-      The shortest route from Salem to Danvers is 3.7 mi:
-            Salem
-            Danvers
-      The shortest route from Salem to Lynn is 4.9 mi:
-            Salem
-            Lynn
-
-------------------------------------------------
-------------------------------------------------
+    The shortest path from Wenham to Hamilton is 1 mi:
+        Wenham
+        Hamilton
 
 The road upgrading goal can be achieved at minimal cost by upgrading:
+    Hamilton to Wenham
 
-      Salem to Beverly
-      Danvers to Beverly
-      Wenham to Danvers
-      Salem to Lynn
-
-------------------------------------------------
-------------------------------------------------
-
+------------------------------------------------------------------
+The input data is:
 
-------------------------------------------------
----------------- New DataSet: ------------------
-------------------------------------------------
+A
+        B 1 mi via bridge
+B
+        A 1 mi via bridge
+        C 1 mi via bridge
+C
+        B 1 mi via bridge
 
-The input data is :
+The shortest paths from A are:
 
-   Wenham
-         Hamilton 1 mi
-   Hamilton
-         Wenham 1 mi
-
-
-The shortest routes from Wenham are:
-
-      The shortest route from Wenham to Hamilton is 1 mi:
-            Wenham
-            Hamilton
-
-------------------------------------------------
-------------------------------------------------
+    The shortest path from A to B is 1 mi:
+        A
+        B
+    The shortest path from A to C is 2 mi:
+        A
+        B
+        C
 
 The road upgrading goal can be achieved at minimal cost by upgrading:
+    B to A
+    C to B
 
-      Hamilton to Wenham
-
-------------------------------------------------
-------------------------------------------------
-
+------------------------------------------------------------------

test-data/shortest.out

@@ -0,0 +1,45 @@
+The input data is:
+
+Wenham
+        Hamilton 1 mi
+Hamilton
+        Wenham 1 mi
+
+The shortest paths from Wenham are:
+
+    The shortest path from Wenham to Hamilton is 1 mi:
+        Wenham
+        Hamilton
+
+The road upgrading goal can be achieved at minimal cost by upgrading:
+    Hamilton to Wenham
+
+
+------------------------------------------------------------------
+
+The input data is:
+
+A
+        B 1 mi via bridge
+B
+        A 1 mi via bridge
+        C 1 mi via bridge
+C
+        B 1 mi via bridge
+
+The shortest paths from A are:
+
+    The shortest path from A to B is 1 mi:
+        A
+        B
+    The shortest path from A to C is 2 mi:
+        A
+        B
+        C
+
+The road upgrading goal can be achieved at minimal cost by upgrading:
+    B to A
+    C to B
+
+
+------------------------------------------------------------------