【update】笔试-滴滴-180918

D-笔试面经/笔试-滴滴-180918.md

@@ -16,10 +16,12 @@ Index
 <div align="center"><img src="../_assets/TIM截图20180918200953.png" height="" /></div>
 
 **思路**
-- DFS
+- DFS（会超时）
+- 多维 DP
+    > [Ways to arrange Balls such that adjacent balls are of different types](https://www.geeksforgeeks.org/ways-to-arrange-balls-such-that-adjacent-balls-are-of-different-types/) - GeeksforGeeks 
 
 **C++**（67%，TLE）
-```
+```C++
 #include <iostream>
 #include <vector>
 
@@ -74,10 +76,160 @@ int main() {
 }
 ```
 
+**多维 DP**（未测试）
+> [Ways to arrange Balls such that adjacent balls are of different types](https://www.geeksforgeeks.org/ways-to-arrange-balls-such-that-adjacent-balls-are-of-different-types/) - GeeksforGeeks 
+
+```C++
+#include<bits/stdc++.h> 
+using namespace std; 
+#define MAX 100 
+  
+// table to store to store results of subproblems 
+int dp[MAX][MAX][MAX][3]; 
+  
+// Returns count of arrangements where last placed ball is 
+// 'last'.  'last' is 0 for 'p', 1 for 'q' and 2 for 'r' 
+int countWays(int p, int q, int r, int last) 
+{ 
+    // if number of balls of any color becomes less 
+    // than 0 the number of ways arrangements is 0. 
+    if (p<0 || q<0 || r<0) 
+        return 0; 
+  
+    // If last ball required is of type P and the number 
+    // of balls of P type is 1 while number of balls of 
+    // other color is 0 the number of ways is 1. 
+    if (p==1 && q==0 && r==0 && last==0) 
+        return 1; 
+  
+    // Same case as above for 'q' and 'r' 
+    if (p==0 && q==1 && r==0 && last==1) 
+        return 1; 
+    if (p==0 && q==0 && r==1 && last==2) 
+        return 1; 
+  
+    // If this subproblem is already evaluated 
+    if (dp[p][q][r][last] != -1) 
+        return dp[p][q][r][last]; 
+  
+    // if last ball required is P and the number of ways is 
+    // the sum of number of ways to form sequence with 'p-1' P 
+    // balls, q Q Balls and r R balls ending with Q and R. 
+    if (last==0) 
+       dp[p][q][r][last] = countWays(p-1,q,r,1) + countWays(p-1,q,r,2); 
+  
+    // Same as above case for 'q' and 'r' 
+    else if (last==1) 
+       dp[p][q][r][last] = countWays(p,q-1,r,0) + countWays(p,q-1,r,2); 
+    else //(last==2) 
+       dp[p][q][r][last] =  countWays(p,q,r-1,0) + countWays(p,q,r-1,1); 
+  
+    return dp[p][q][r][last]; 
+} 
+  
+// Returns count of required arrangements 
+int countUtil(int p, int q, int r) 
+{ 
+   // Initialize 'dp' array 
+   memset(dp, -1, sizeof(dp)); 
+  
+   // Three cases arise: 
+   return countWays(p, q, r, 0) +  // Last required balls is type P 
+          countWays(p, q, r, 1) +  // Last required balls is type Q 
+          countWays(p, q, r, 2); // Last required balls is type R 
+} 
+  
+// Driver code to test above 
+int main() 
+{ 
+    int p = 1, q = 1, r = 1; 
+    printf("%d", countUtil(p, q, r)); 
+    return 0; 
+} 
+```
+
 ## 交通轨迹分析
 <div align="center"><img src="../_assets/TIM截图20180918201038.png" height="" /></div>
 <div align="center"><img src="../_assets/TIM截图20180918201054.png" height="" /></div>
 
 **思路**
-- 并查集
-- 判断相交
+- 并查集/暴力枚举
+- 判断相交
+    ```python
+    class Point(object):
+        def __init__(self, x, y):
+            self.x = x
+            self.y = y
+
+
+    def is_cross(a, b, c, d):
+        fc = (c.y - a.y) * (a.x - b.x) - (c.x - a.x) * (a.y - b.y)
+        fd = (d.y - a.y) * (a.x - b.x) - (d.x - a.x) * (a.y - b.y)
+
+        if fc * fd > 0:
+            return False
+        return True
+    ```
+
+**暴力**（未测试）
+```python
+# 作者：牛客4807725号
+# 链接：https://www.nowcoder.com/discuss/112681?toCommentId=1905196
+# 来源：牛客网
+
+class point():
+    def __init__(self, x, y):
+        self.x = x
+        self.y = y
+
+
+def isxiangjiao(a, b, c, d):
+    fc = (c.y - a.y) * (a.x - b.x) - (c.x - a.x) * (a.y - b.y)
+    fd = (d.y - a.y) * (a.x - b.x) - (d.x - a.x) * (a.y - b.y)
+    if fc * fd > 0:
+        return False
+    return True
+
+
+import sys
+import collections
+
+if __name__ == "__main__":
+    t = int(sys.stdin.readline().strip())
+    for i in range(t):
+        n = int(sys.stdin.readline().strip())
+        roaddict = collections.defaultdict(set)
+        roadpos = []
+        count = 1
+        for j in range(n):
+            line = sys.stdin.readline().strip().split()
+            if line[0] == 'T':
+                x1, y1, x2, y2 = int(line[1]), int(line[2]), int(line[3]), int(line[4])
+                pointa = point(x1, y1)
+                pointb = point(x2, y2)
+                roadpos.append([pointa, pointb])
+                if count == 1:
+                    roaddict[count].add(count)
+                    count += 1
+                else:
+                    visited = set()
+                    for tmpi in range(1, len(roadpos) + 1):
+                        pointc = roadpos[tmpi - 1][0]
+                        pointd = roadpos[tmpi - 1][1]
+                        if tmpi not in visited and isxiangjiao(pointa, pointb, pointc, pointd):
+                            visited.add(tmpi)
+                            roaddict[tmpi].add(count)
+                            roaddict[count].add(count)
+                            for tmp in roaddict[tmpi]:
+                                visited.add(tmp)
+                                roaddict[tmp].add(count)
+                                roaddict[count].add(tmp)
+                    for tmp in roaddict[count]:
+                        roaddict[tmp] = roaddict[count]
+                    count += 1
+            if line[0] == 'Q':
+                #                 print(roaddict)
+                queryvale = int(line[1])
+                print(len(roaddict[queryvale]))
+        print()
+```