题目
给定一个无向图，两个顶点之间可能有多条边，求任意两个顶点之间的最短路径。
具体描述请见hihoCoder。
解题思路
经典的Floyd算法。采用动态规划的思想，假设d[i,
j]是从i到j的距离，且路径允许经过的节点数慢慢增加。写成递推式就是
d[i, j] = min(d[i, j], d[i, k] + d[k, j])
时间复杂度
时间复杂度为N3，空间复杂度为N2。
代码
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| #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
void floyd(int** graph, int N) {
for (int k = 1; k <= N; k++) {
for (int i = 1; i <= N; i++)
for (int j = 1; j <= N; j++) {
if (graph[i][k] != -1 && graph[k][j] != -1) {
if (graph[i][j] == -1)
graph[i][j] = graph[i][k] + graph[k][j];
else
graph[i][j] = min(graph[i][j], graph[i][k] + graph[k][j]);
}
}
}
}
int main() {
int N = 0, M = 0;
cin >> N >> M;
int** graph = new int*[N + 1];
for (int i = 1; i < N + 1; i++) {
graph[i] = new int[N + 1];
for (int j = 1; j < N + 1; j++)
graph[i][j] = -1;
graph[i][i] = 0;
}
for (int i = 0; i < M; i++) {
int a = 0, b = 0, edge = 0;
cin >> a >> b >> edge;
if (graph[a][b] == -1) {
graph[a][b] = edge;
graph[b][a] = edge;
}
else if (graph[a][b] > edge) {
graph[a][b] = edge;
graph[b][a] = edge;
}
}
floyd(graph, N);
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= N; j++)
cout << graph[i][j] << " ";
cout << endl;
}
for (int i = 1; i < N + 1; i++)
delete[] graph[i];
delete[] graph;
return 0;
}
|
