A-M baby

Task/Knapsack-problem-Unbounded/Fortran/knapsack-problem-unbounded.f

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+PROGRAM KNAPSACK
+
+  IMPLICIT NONE
+
+  REAL :: totalWeight, totalVolume
+  INTEGER :: maxPanacea, maxIchor, maxGold, maxValue = 0
+  INTEGER :: i, j, k
+  INTEGER :: n(3)
+
+  TYPE Bounty
+    INTEGER :: value
+    REAL :: weight
+    REAL :: volume
+  END TYPE Bounty
+
+  TYPE(Bounty) :: panacea, ichor, gold, sack, current
+
+  panacea = Bounty(3000, 0.3, 0.025)
+  ichor   = Bounty(1800, 0.2, 0.015)
+  gold    = Bounty(2500, 2.0, 0.002)
+  sack    = Bounty(0, 25.0, 0.25)
+
+  maxPanacea = MIN(sack%weight / panacea%weight, sack%volume / panacea%volume)
+  maxIchor = MIN(sack%weight / ichor%weight, sack%volume / ichor%volume)
+  maxGold = MIN(sack%weight / gold%weight, sack%volume / gold%volume)
+
+  DO i = 0, maxPanacea
+     DO j = 0, maxIchor
+        Do k = 0, maxGold
+           current%value = k * gold%value + j * ichor%value + i * panacea%value
+           current%weight = k * gold%weight + j * ichor%weight + i * panacea%weight
+           current%volume = k * gold%volume + j * ichor%volume + i * panacea%volume
+           IF (current%weight > sack%weight .OR. current%volume > sack%volume) CYCLE
+           IF (current%value > maxValue) THEN
+              maxValue = current%value
+              totalWeight = current%weight
+              totalVolume = current%volume
+              n(1) = i ; n(2) = j ; n(3) = k
+           END IF
+        END DO
+     END DO
+  END DO
+
+  WRITE(*, "(A,I0)") "Maximum value achievable is ", maxValue
+  WRITE(*, "(3(A,I0),A)") "This is achieved by carrying ", n(1), " panacea, ", n(2), " ichor and ", n(3), " gold items"
+  WRITE(*, "(A,F4.1,A,F5.3)") "The weight to carry is ", totalWeight, " and the volume used is ", totalVolume
+
+END PROGRAM KNAPSACK