题目描述
用邻接矩阵存储有向图，实现最短路径Dijkstra算法，图中边的权值为整型，顶点个数少于10个。
部分代码提示：
#include
#include
using namespace std;
const int MaxSize = 10;
const int INF = 32767;
class MGraph
{
public:
MGraph(char a[], int n, int e);
void Dijkstra();
private:
char vertex[MaxSize];
int arc[MaxSize][MaxSize];
int vertexNum, arcNum;
};
MGraph::MGraph(char a[], int n, int e)
{
//write your code.
}
int Min(int dist[], int vertexNum)
{
//write your code.
}
void MGraph::Dijkstra()
{
//write your code.
}
int main()
{
int n = 0;
int e = 0;
cin >> n >> e;
char p[MaxSize];
int i = 0;
for (i=0; i<n; i++)
{cin >> p[i];
}MGraph MG(p, n, e);MG.Dijkstra();return 0;
}
输入描述
首先输入图中顶点个数和边的条数；
再输入顶点的信息（字符型）；
再输入各边及其权值。
输出描述
依次输出从编号为0的顶点开始的从小到大的所有最短路径，每条路径及其长度占一行。
输入样例
5 7
A B C D E
0 1 6
0 2 2
0 3 1
1 2 4
1 3 3
2 4 6
3 4 7
输出样例
AD 1
AC 2
AB 6
ADE 8
内存阀值:50240K 耗时阀值:5000MS
代码
#include <iostream>
#include <string>
using namespace std;const int MaxSize = 10;
const int INF = 32767;class MGraph
{
public:MGraph(string a[], int n, int e);void Dijkstra();private:string vertex[MaxSize];int arc[MaxSize][MaxSize];int vertexNum, arcNum;
};MGraph::MGraph(string a[], int n, int e)
{//1、赋值vertexNum = n;arcNum = e;//2、顶点表for (int i = 0; i < vertexNum; i++)vertex[i] = a[i];//3、边表初始化for (int i = 0; i < vertexNum; i++)for (int j = 0; j < vertexNum; j++)arc[i][j] = INF;//4、边表赋值int i = 0, j = 0, w = 0;for (int k = 0; k < arcNum; k++){cin >> i >> j >> w;arc[i][j] = w;//arc[j][i] = w;//有方向的不用反复设置相同！！！！掉坑了我淦！}
}int Min(int dist[], int vertexNum)
{int min = INF;int pos = 0;for (int i = 0; i < vertexNum; i++){if (dist[i] < min && dist[i] != 0)//!=0?:存入的顶点不需要遍历{min = dist[i];pos = i;}}return pos;
}void MGraph::Dijkstra()
{int v = 0;int i, k, num, dist[MaxSize];//权值表string path[MaxSize];//路径表for (i = 0; i < vertexNum; i++){dist[i] = arc[v][i];//v-表示当前顶点，i-表示这个顶点与其他顶点的边的权值（希望屏幕前的你能理解）if (dist[i] != INF)//如果连通path[i] = vertex[v] + vertex[i];elsepath[i] = "";}dist[0] = 0;for (num = 1; num < vertexNum; num++)//不知道为什么=1？顶点表已经在集合中了，所以不用遍历{k = Min(dist, vertexNum);cout << path[k] <<' ' << dist[k];for (i = 0; i < vertexNum; i++)//更新最小权值表{if (dist[i] > dist[k] + arc[k][i]){dist[i] = dist[k] + arc[k][i];path[i] = path[k] + vertex[i];}}dist[k] = 0;//将k加入集合，为什么是0？因为0比其他路径权值和都要小所以无法改变}
}int main()
{int n = 0;int e = 0;cin >> n >> e;string p[MaxSize];int i = 0;for (i = 0; i < n; i++){cin >> p[i];}MGraph MG(p, n, e);MG.Dijkstra();return 0;
}
