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yanglbmeyanglbme
committedJan 24, 2022
feat: add solutions to lc problem: No.2045
No.2045.Second Minimum Time to Reach Destination
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‎solution/1700-1799/1705.Maximum Number of Eaten Apples/README.md

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Original file line numberDiff line numberDiff line change
@@ -113,7 +113,7 @@ class Solution {
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### **C++**
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```cpp
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typedef pair<int,int> PII;
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typedef pair<int, int> PII;
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class Solution {
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public:

‎solution/1700-1799/1705.Maximum Number of Eaten Apples/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -103,7 +103,7 @@ class Solution {
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### **C++**
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```cpp
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typedef pair<int,int> PII;
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typedef pair<int, int> PII;
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class Solution {
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public:

‎solution/1700-1799/1705.Maximum Number of Eaten Apples/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -1,4 +1,4 @@
1-
typedef pair<int,int> PII;
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typedef pair<int, int> PII;
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class Solution {
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public:

‎solution/2000-2099/2045.Second Minimum Time to Reach Destination/README.md

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Original file line numberDiff line numberDiff line change
@@ -89,15 +89,83 @@
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def secondMinimum(self, n: int, edges: List[List[int]], time: int, change: int) -> int:
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g = defaultdict(set)
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for u, v in edges:
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g[u].add(v)
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g[v].add(u)
98+
q = deque([(1, 0)])
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dist = [[float('inf')] * 2 for _ in range(n + 1)]
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dist[1][1] = 0
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while q:
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u, d = q.popleft()
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for v in g[u]:
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if d + 1 < dist[v][0]:
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dist[v][0] = d + 1
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q.append((v, d + 1))
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elif dist[v][0] < d + 1 < dist[v][1]:
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dist[v][1] = d + 1
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if v == n:
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break
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q.append((v, d + 1))
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ans = 0
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for i in range(dist[n][1]):
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ans += time
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if i < dist[n][1] - 1 and (ans // change) % 2 == 1:
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ans = (ans + change) // change * change
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return ans
93118
```
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### **Java**
96121

97122
<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
100-
125+
class Solution {
126+
public int secondMinimum(int n, int[][] edges, int time, int change) {
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List<Integer>[] g = new List[n + 1];
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for (int i = 0; i < n + 1; ++i) {
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g[i] = new ArrayList<>();
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}
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for (int[] e : edges) {
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int u = e[0], v = e[1];
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g[u].add(v);
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g[v].add(u);
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}
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Deque<int[]> q = new LinkedList<>();
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q.offerLast(new int[]{1, 0});
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int[][] dist = new int[n + 1][2];
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for (int i = 0; i < n + 1; ++i) {
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Arrays.fill(dist[i], Integer.MAX_VALUE);
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}
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dist[1][1] = 0;
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while (!q.isEmpty()) {
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int[] e = q.pollFirst();
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int u = e[0], d = e[1];
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for (int v : g[u]) {
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if (d + 1 < dist[v][0]) {
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dist[v][0] = d + 1;
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q.offerLast(new int[]{v, d + 1});
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} else if (dist[v][0] < d + 1 && d + 1 < dist[v][1]) {
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dist[v][1] = d + 1;
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if (v == n) {
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break;
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}
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q.offerLast(new int[]{v, d + 1});
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}
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}
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}
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int ans = 0;
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for (int i = 0; i < dist[n][1]; ++i) {
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ans += time;
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if (i < dist[n][1] - 1 && (ans / change) % 2 == 1) {
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ans = (ans + change) / change * change;
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}
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}
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return ans;
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}
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}
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```
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103171
### **...**

‎solution/2000-2099/2045.Second Minimum Time to Reach Destination/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -78,13 +78,81 @@ The second minimum time path is 1 -&gt; 2 -&gt; 1 -&gt; 2 with time = 11 minutes
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### **Python3**
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```python
81-
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class Solution:
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def secondMinimum(self, n: int, edges: List[List[int]], time: int, change: int) -> int:
83+
g = defaultdict(set)
84+
for u, v in edges:
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g[u].add(v)
86+
g[v].add(u)
87+
q = deque([(1, 0)])
88+
dist = [[float('inf')] * 2 for _ in range(n + 1)]
89+
dist[1][1] = 0
90+
while q:
91+
u, d = q.popleft()
92+
for v in g[u]:
93+
if d + 1 < dist[v][0]:
94+
dist[v][0] = d + 1
95+
q.append((v, d + 1))
96+
elif dist[v][0] < d + 1 < dist[v][1]:
97+
dist[v][1] = d + 1
98+
if v == n:
99+
break
100+
q.append((v, d + 1))
101+
ans = 0
102+
for i in range(dist[n][1]):
103+
ans += time
104+
if i < dist[n][1] - 1 and (ans // change) % 2 == 1:
105+
ans = (ans + change) // change * change
106+
return ans
82107
```
83108

84109
### **Java**
85110

86111
```java
87-
112+
class Solution {
113+
public int secondMinimum(int n, int[][] edges, int time, int change) {
114+
List<Integer>[] g = new List[n + 1];
115+
for (int i = 0; i < n + 1; ++i) {
116+
g[i] = new ArrayList<>();
117+
}
118+
for (int[] e : edges) {
119+
int u = e[0], v = e[1];
120+
g[u].add(v);
121+
g[v].add(u);
122+
}
123+
Deque<int[]> q = new LinkedList<>();
124+
q.offerLast(new int[]{1, 0});
125+
int[][] dist = new int[n + 1][2];
126+
for (int i = 0; i < n + 1; ++i) {
127+
Arrays.fill(dist[i], Integer.MAX_VALUE);
128+
}
129+
dist[1][1] = 0;
130+
while (!q.isEmpty()) {
131+
int[] e = q.pollFirst();
132+
int u = e[0], d = e[1];
133+
for (int v : g[u]) {
134+
if (d + 1 < dist[v][0]) {
135+
dist[v][0] = d + 1;
136+
q.offerLast(new int[]{v, d + 1});
137+
} else if (dist[v][0] < d + 1 && d + 1 < dist[v][1]) {
138+
dist[v][1] = d + 1;
139+
if (v == n) {
140+
break;
141+
}
142+
q.offerLast(new int[]{v, d + 1});
143+
}
144+
}
145+
}
146+
int ans = 0;
147+
for (int i = 0; i < dist[n][1]; ++i) {
148+
ans += time;
149+
if (i < dist[n][1] - 1 && (ans / change) % 2 == 1) {
150+
ans = (ans + change) / change * change;
151+
}
152+
}
153+
return ans;
154+
}
155+
}
88156
```
89157

90158
### **...**

‎solution/2000-2099/2045.Second Minimum Time to Reach Destination/Solution.java

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,44 @@
1+
class Solution {
2+
public int secondMinimum(int n, int[][] edges, int time, int change) {
3+
List<Integer>[] g = new List[n + 1];
4+
for (int i = 0; i < n + 1; ++i) {
5+
g[i] = new ArrayList<>();
6+
}
7+
for (int[] e : edges) {
8+
int u = e[0], v = e[1];
9+
g[u].add(v);
10+
g[v].add(u);
11+
}
12+
Deque<int[]> q = new LinkedList<>();
13+
q.offerLast(new int[]{1, 0});
14+
int[][] dist = new int[n + 1][2];
15+
for (int i = 0; i < n + 1; ++i) {
16+
Arrays.fill(dist[i], Integer.MAX_VALUE);
17+
}
18+
dist[1][1] = 0;
19+
while (!q.isEmpty()) {
20+
int[] e = q.pollFirst();
21+
int u = e[0], d = e[1];
22+
for (int v : g[u]) {
23+
if (d + 1 < dist[v][0]) {
24+
dist[v][0] = d + 1;
25+
q.offerLast(new int[]{v, d + 1});
26+
} else if (dist[v][0] < d + 1 && d + 1 < dist[v][1]) {
27+
dist[v][1] = d + 1;
28+
if (v == n) {
29+
break;
30+
}
31+
q.offerLast(new int[]{v, d + 1});
32+
}
33+
}
34+
}
35+
int ans = 0;
36+
for (int i = 0; i < dist[n][1]; ++i) {
37+
ans += time;
38+
if (i < dist[n][1] - 1 && (ans / change) % 2 == 1) {
39+
ans = (ans + change) / change * change;
40+
}
41+
}
42+
return ans;
43+
}
44+
}

‎solution/2000-2099/2045.Second Minimum Time to Reach Destination/Solution.py

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@@ -0,0 +1,26 @@
1+
class Solution:
2+
def secondMinimum(self, n: int, edges: List[List[int]], time: int, change: int) -> int:
3+
g = defaultdict(set)
4+
for u, v in edges:
5+
g[u].add(v)
6+
g[v].add(u)
7+
q = deque([(1, 0)])
8+
dist = [[float('inf')] * 2 for _ in range(n + 1)]
9+
dist[1][1] = 0
10+
while q:
11+
u, d = q.popleft()
12+
for v in g[u]:
13+
if d + 1 < dist[v][0]:
14+
dist[v][0] = d + 1
15+
q.append((v, d + 1))
16+
elif dist[v][0] < d + 1 < dist[v][1]:
17+
dist[v][1] = d + 1
18+
if v == n:
19+
break
20+
q.append((v, d + 1))
21+
ans = 0
22+
for i in range(dist[n][1]):
23+
ans += time
24+
if i < dist[n][1] - 1 and (ans // change) % 2 == 1:
25+
ans = (ans + change) // change * change
26+
return ans

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