Codejam to I/O for Women 2017

Author: shixi_chenzhang

IO for Women/2017/A.java

@@ -1,6 +1,10 @@
 import java.io.PrintStream;
 import java.util.*;
 
+/**
+ * Codejam to I/O for Women 2017 Problem A: Ticket Trouble
+ * Check README.md for explanation.
+ */
 public class Main {
 
     private String solve(Scanner scanner) {

IO for Women/2017/B.java

@@ -1,6 +1,10 @@
 import java.io.PrintStream;
 import java.util.*;
 
+/**
+ * Codejam to I/O for Women 2017 Problem B: Understudies
+ * Check README.md for explanation.
+ */
 public class Main {
 
     private String solve(Scanner scanner) {

IO for Women/2017/C.java

@@ -1,6 +1,10 @@
 import java.io.PrintStream;
 import java.util.*;
 
+/**
+ * Codejam to I/O for Women 2017 Problem C: Word Search
+ * Check README.md for explanation.
+ */
 public class Main {
 
     private String solve(Scanner scanner) {

IO for Women/2017/D.java

@@ -1,6 +1,10 @@
 import java.io.PrintStream;
 import java.util.*;
 
+/**
+ * Codejam to I/O for Women 2017 Problem D: Where Ya Gonna Call?
+ * Check README.md for explanation.
+ */
 public class Main {
 
     private String solve(Scanner scanner) {

IO for Women/2017/README.md

@@ -0,0 +1,39 @@
+# Codejam to I/O for Women 2017
+https://code.google.com/codejam/contest/12224486/dashboard
+
+## Problem A: Ticket Trouble
+
+Simple. For each row, greedy to get the maximum number.
+
+## Problem B: Understudies
+
+Sort all probabilities. Then match the smallest one with the biggest one.
+
+## Problem C: Word Search
+
+Not so simple.  
+Just consider of this pattern:
+```
+IIIIIIIIIIIIII
+//////////////
+OOOOOOOOOOOOOO
+//////////////
+IIIIIIIIIIIIII
+...
+```
+For each `/` on border, it will create 1 `I/O`.  
+For each `/` in middle, it will create 3 `I/O`.
+
+## Problem D: Where Ya Gonna Call?
+
+This problem is hard, very hard, maybe the hardest problem in Codejam to I/O for Women.  
+Using Floyd algorithm to calculate distance between any nodes.  
+Then, for each edge, for each possible position, check the minimax.  
+The problem is, what's the possible position?  
+
+Assume the verticals of the edge is `u, v` and its length is `l`.  
+Assume the distance between u and any point is `a[]`, distance
+between v and any point is `b[]`.
+Then, all possible positions x should satisfies `a[i]+x = b[j]+(l-x)`.  
+So, just iterate i and j, to calculate all possible x, and find the minimax.
+