目录
一、题目
1、题目描述
给你一个有
n个节点的 有向带权 图，节点编号为0到n - 1。图中的初始边用数组edges表示，其中edges[i] = [fromi, toi, edgeCosti]表示从fromi到toi有一条代价为edgeCosti的边。请你实现一个
Graph类：
Graph(int n, int[][] edges)初始化图有n个节点，并输入初始边。addEdge(int[] edge)向边集中添加一条边，其中edge = [from, to, edgeCost]。数据保证添加这条边之前对应的两个节点之间没有有向边。int shortestPath(int node1, int node2)返回从节点node1到node2的路径最小 代价。如果路径不存在，返回-1。一条路径的代价是路径中所有边代价之和。
2、接口描述
cpp
cpp
class Graph {
public:
Graph(int n, vector<vector<int>>& edges) {
}
void addEdge(vector<int> edge) {
}
int shortestPath(int node1, int node2) {
}
};
/**
* Your Graph object will be instantiated and called as such:
* Graph* obj = new Graph(n, edges);
* obj->addEdge(edge);
* int param_2 = obj->shortestPath(node1,node2);
*/python3
cpp
class Graph:
def __init__(self, n: int, edges: List[List[int]]):
def addEdge(self, edge: List[int]) -> None:
def shortestPath(self, node1: int, node2: int) -> int:
# Your Graph object will be instantiated and called as such:
# obj = Graph(n, edges)
# obj.addEdge(edge)
# param_2 = obj.shortestPath(node1,node2)3、原题链接
2642. 设计可以求最短路径的图类 - 力扣（LeetCode）
二、解题报告
1、思路分析
数据规模100个点，可以用floyd
我们初始化就直接floyd即可
然后考虑floyd的原理无非就是枚举路径上的点进行动态规划，那么我们对<x, y>这条边进行修改后，以<x, y>这条边进行一次松弛操作即可
2、复杂度
时间复杂度：初始化：O(n^3) ，最短路维护：O(n^2)，最短路获取：O(1)空间复杂度：O(n^2)
3、代码详解
cpp
cpp
int g[105][105];
const int inf = 0x3f3f3f3f;
class Graph {
public:
int _n;
Graph(int n, vector<vector<int>>& edges):_n(n) {
memset(g, 0x3f, sizeof g);
for(int i = 0; i < n; i++) g[i][i] = 0;
for(auto& e : edges)
g[e[0]][e[1]] = e[2];
for(int k = 0; k < n; k++)
for(int i = 0; i < n; i++){
if(g[i][k] == inf) continue;
for(int j = 0; j < n; j++)
g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
}
}
void addEdge(vector<int> e) {
if(e[2] >= g[e[0]][e[1]]) return;
int a = e[0], b = e[1];
g[a][b] = e[2];
for(int i = 0; i < _n; i++)
for(int j = 0; j < _n; j++)
g[i][j] = min(g[i][j], g[i][a] + g[b][j] + e[2]);
}
int shortestPath(int x, int y) {
return g[x][y] < inf ? g[x][y] : -1;
}
};
/**
* Your Graph object will be instantiated and called as such:
* Graph* obj = new Graph(n, edges);
* obj->addEdge(edge);
* int param_2 = obj->shortestPath(node1,node2);
*/python3
python
class Graph:
def __init__(self, n: int, edges: List[List[int]]):
self.g = [[inf] * n for _ in range(n)]
g = self.g
for i in range(n):
g[i][i] = 0
for x, y, w in edges:
g[x][y] = w
for k in range(n):
for i in range(n):
for j in range(n):
g[i][j] = min(g[i][j], g[i][k] + g[k][j])
def addEdge(self, edge: List[int]) -> None:
x, y, w = edge
g = self.g
if w > g[x][y]:
return
g[x][y] = w
n = len(self.g)
for i in range(n):
for j in range(n):
g[i][j] = min(g[i][j], g[i][x] + g[y][j] + g[x][y])
def shortestPath(self, x: int, y: int) -> int:
g = self.g
return g[x][y] if g[x][y] < inf else -1
# Your Graph object will be instantiated and called as such:
# obj = Graph(n, edges)
# obj.addEdge(edge)
# param_2 = obj.shortestPath(node1,node2)