책, 백준 숙제

cmkim/BaekJoon/백준_1939.py

@@ -0,0 +1,24 @@
+n, m = map(int, input().split())
+
+arr = [[] for _ in range(n+1)]
+
+for i in range(m):
+    a, b, c = map(int, input().split())
+    arr[a].append([b, c])
+    arr[b].append([a, b])
+
+for i in range(1, n+1):
+    print(arr[i])
+
+start, end = map(int, input().split())
+
+
+# 플로이드 워셜 반대로?
+# for m in range(1, n+1):
+#     for s in range(1, n+1):
+#         for e in range(1, n+1):
+#             if d[s][e] > d[s][m] + d[m][e]:
+#                 d[s][e] = d[s][m] + d[m][e]
+
+
+#print(d)

cmkim/BaekJoon/백준_5972.py

@@ -0,0 +1,28 @@
+import sys
+import heapq
+INF = 1e9
+n, m = map(int, input().split())
+arr = [[] * (n+1) for _ in range(n+1)]
+dis = [INF] * (n+1)
+
+for _ in range(m):
+    a, b, c = map(int, input().split())
+    arr[a].append((b, c))
+    arr[b].append((a, c))
+
+def dijkstra(start):
+    q = []
+    heapq.heappush(q, (0, start)) #비용, 출발점
+    dis[start] = 0
+    while q:
+        d, now = heapq.heappop(q)
+        if dis[now] < d:
+            continue
+        for v, w in arr[now]: # 가려는 곳에대한 위치, 비용
+            cost = d + w
+            if cost < dis[v]:
+                dis[v] = cost
+                heapq.heappush(q, (cost, v))
+
+dijkstra(1)
+print(dis[n])

cmkim/Programmers/스택_큐/다리를 지나는 트럭.py

@@ -1,13 +1,23 @@
 def solution(bridge_length, weight, truck_weights):
+    time = 0
+    on_bridge = []
+    on_time = []
 
-    answer = 0
+    while(truck_weights or on_bridge) :
+        time += 1
 
+        if(on_time) :
+            if(on_time[0] + bridge_length == time) :
+                on_bridge.pop(0)
+                on_time.pop(0)
+        if(truck_weights) :
+            if(sum(on_bridge) + truck_weights[0] <= weight) :
+                on_bridge.append(truck_weights.pop(0))
+                on_time.append(time)
 
-
-    return answer
-
+    return time
 '''
 트럭 무게를 정렬할 필요는 없고
 그냥 순서대로 
+아 무조건 정해진 순서대로 가야한다..
 '''
-

cmkim/이것이 취업을 위한 코딩 테스트다/뱀_327p.py

@@ -0,0 +1,91 @@
+n = int(input())
+k = int(input())
+
+arr = [[0]*n for _ in range(n+1)]
+order = []
+count = 0
+dir = 0
+
+for i in range(k):
+    a, b = map(int, input().split())
+    arr[a][b] = 1
+
+l = int(input())
+
+for i in range(l):
+    # x = int(input().split())
+    # c = input()
+    x, c = input().split()
+    order.append((int(x), c))
+
+# for i in range(n):
+#     print(arr[i])
+# print(order)
+# order.pop()
+#print(order)
+
+#오른쪽 볼때 좌, 우 회전 순서는 각각 우 상 좌 하 / 우 하 좌 상
+#왼쪽 ''                           좌 하 우 상 / 좌 상 우 하
+
+dx = [0, -1, 0, 1] #우 상 좌 하
+dy = [1, 0, -1, 0]
+
+x, y = 1, 1
+q = [(x, y)]
+
+while True:
+    count += 1
+
+    arr[x][y] = 2
+
+
+
+    nx = x + dx[dir]
+    ny = y + dy[dir]
+
+    if 1 <= nx <= n and 1 <= ny <= n and arr[nx][ny] != 2:
+        if arr[nx][ny] == 0:
+            arr[nx][ny] = 2
+            q.append((nx, ny))
+            ex, ey = q.pop(0)
+            arr[ex][ey] = 0
+
+        elif arr[nx][ny] == 1:
+            arr[nx][ny] = 2
+            q.append((nx, ny))
+
+        x, y = nx, ny
+
+    else:
+        break
+
+    if count == order[0][0]:
+        if order[0][1] == 'L': #왼쪽회전
+            dir = (dir+1) % 4
+        else:                  #오른쪽회전
+            dir = (dir+3) % 4
+
+        order.pop()
+
+print(count)
+
+
+
+
+
+
+
+
+
+
+'''
+6
+3
+3 4
+2 5
+5 3
+3
+3 D
+15 L
+17  D
+'''

cmkim/이것이 취업을 위한 코딩 테스트다/치킨 배달_332p.py

@@ -0,0 +1,41 @@
+from itertools import combinations
+n, m = map(int, input().split())
+
+arr = []
+chicken = []
+home = []
+
+for i in range(n):
+    temp = list(map(int, input().split()))
+    for j in range(n):
+        if temp[j] == 1:
+            home.append((i, j))
+        elif temp[j] == 2:
+            chicken.append((i, j))
+
+m_chicken = list(combinations(chicken, m))
+
+def get_sum(candidate):
+    result = 0
+    for hx, hy in home:
+        temp = 1e9
+        for cx, cy in candidate:
+            temp = min(temp, abs(hx - cx) + abs(hy - cy))
+        result += temp
+    return result
+
+result = 1e9
+for i in m_chicken:
+    result = min(result, get_sum(i))
+
+
+print(result)
+
+'''
+5 3
+0 0 1 0 0
+0 0 2 0 1
+0 1 2 0 0
+0 0 1 0 0
+0 0 0 0 2
+'''