EXY-SC-1380
第 121 题
下面的程序使用出边邻接表表达的带权无向图,则从顶点 0 到顶点 3 的最短距离为( )。
#include <vector>
using namespace std;
class Edge {
public:
int dest;
int weight;
Edge(int d, int w) : dest(d), weight(w) {}
};
class Graph {
private:
int num_vertex;
vector<vector<Edge>> vve;
public:
Graph(int v) : num_vertex(v), vve(v) {}
void addEdge(int s, int d, int w) {
vve[s].emplace_back(d, w);
vve[d].emplace_back(s, w);
}
};
int main() {
Graph g(4);
g.addEdge(0, 1, 8);
g.addEdge(0, 2, 5);
g.addEdge(1, 2, 1);
g.addEdge(1, 3, 3);
g.addEdge(2, 3, 7);
return 0;
}
语言:
C++
GESP真题
八级
2025.9
单选题号:
15
EXY-SC-1379
第 122 题
下面 Prim 算法程序中,横线处应该填入的是( )。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int prim(vector<vector<int>> & graph, int n) {
vector<int> key(n, INT_MAX);
vector<int> parent(n, -1);
key[0] = 0;
for (int i = 0; i < n; i++) {
int u = min_element(key.begin(), key.end()) - key.begin();
if (key[u] == INT_MAX)
break;
for (int v = 0; v < n; v++) {
if (__________) { // 在此处填入选项
key[v] = graph[u][v];
parent[v] = u;
}
}
}
int sum = 0;
for (int i = 0; i < n; i++) {
if (parent[i] != -1) {
cout << "Edge: " << parent[i] << " - " << i << " Weight: " << key[i] << endl;
sum += key[i];
}
}
return sum;
}
int main() {
int n, m;
cin >> n >> m;
vector<vector<int>> graph(n, vector<int>(n, 0));
for (int i = 0; i < m; i++) {
int u, v, w;
cin >> u >> v >> w;
graph[u][v] = w;
graph[v][u] = w;
}
int result = prim(graph, n);
cout << "Total weight of the minimum spanning tree: " << result << endl;
return 0;
}
语言:
C++
GESP真题
八级
2025.9
单选题号:
14
EXY-SC-1378
第 123 题
下面 merge_sort 函数试图实现归并排序算法,横线处应该填入的是( )。
#include <vector>
using namespace std;
void merge_sort(vector<int> & arr, int left, int right) {
if (right - left <= 1)
return;
int mid = (left + right) / 2;
merge_sort(__________); // 在此处填入选项
merge_sort(__________); // 在此处填入选项
vector<int> temp(right - left);
int i = left, j = mid, k = 0;
while (i < mid && j < right)
if (arr[i] <= arr[j])
temp[k++] = arr[i++];
else
temp[k++] = arr[j++];
while (i < mid)
temp[k++] = arr[i++];
while (j < right)
temp[k++] = arr[j++];
for (i = left, k = 0; i < right; ++i, ++k)
arr[i] = temp[k];
}
语言:
C++
GESP真题
八级
2025.9
单选题号:
13
EXY-SC-1377
第 124 题
下面 count_triple 函数的时间复杂度为( )。
int gcd(int m, int n) {
if (m == 0) return n;
return gcd(n % m, m);
}
int count_triple(int n) {
int cnt = 0;
for (int v = 1; v * v * 4 <= n; v++)
for (int u = v + 1; u * (u + v) * 2 <= n; u += 2)
if (gcd(u, v) == 1) {
int a = u * u - v * v;
int b = u * v * 2;
int c = u * u + v * v;
cnt += n / (a + b + c);
}
return cnt;
}
语言:
C++
GESP真题
八级
2025.9
单选题号:
12
EXY-SC-1376
第 125 题
下列 Dijkstra 算法,假设图 graph 中顶点数 $v$、边数 $e$,则程序的时间复杂度为( )。
typedef struct Edge {
int in, out; // 从下标in顶点到下标out顶点的边
int len; // 边长度
struct Edge * next;
} Edge;
// v: 顶点个数, graph: 出边邻接表, start: 起点下标, dis: 输出每个顶点的最短距离
void dijkstra(int v, Edge * graph[], int start, int * dis) {
const int MAX_DIS = 0x7fffffff;
for (int i = 0; i < v; i++)
dis[i] = MAX_DIS;
dis[start] = 0;
int * visited = new int[v];
for (int i = 0; i < v; i++)
visited[i] = 0;
visited[start] = 1;
for (int t = 0; ; t++) {
int min = MAX_DIS, minv = -1;
for (int i = 0; i < v; i++) {
if (visited[i] == 0 && min > dis[i]) {
min = dis[i];
minv = i;
}
}
if (minv < 0)
break;
visited[minv] = 1;
for (Edge * e = graph[minv]; e != NULL; e = e->next)
if (dis[e->out] > e->len)
dis[e->out] = e->len;
}
delete[] visited;
}
语言:
C++
GESP真题
八级
2025.9
单选题号:
11
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