SDUT 2143 图结构练习——最短路径(Dijkstra & Floyed)

图结构练习——最短路径

Time Limit: 1000 ms Memory Limit: 65536 KiB

Submit Statistic

Problem Description

 给定一个带权无向图，求节点1到节点n的最短路径。

 

Input

 输入包含多组数据，格式如下。

第一行包括两个整数n m，代表节点个数和边的个数。(n<=100)

剩下m行每行3个正整数a b c，代表节点a和节点b之间有一条边，权值为c。

 

Output

 每组输出占一行，仅输出从1到n的最短路径权值。（保证最短路径存在）

 

Sample Input

3 2
1 2 1
1 3 1
1 0

Sample Output

1
0

Hint

 

Source

赵利强

#include 
using namespace std;
#define INF INT_MAX

int n, m;
int gra[111][111], cat[111], dist[111];

void Dijkstra(int k)
{
    for (int i = 1; i <= n; i++) // dist数组初始化
        dist[i] = gra[k][i];

    dist[k] = 0; // 初始化起点
    cat[k] = 1;

    for (int q = 1; q <= n; q++)
    {
        int min = INF;
        int index = k; // 记录下一个确定点的序号
        for (int j = 1; j <= n; j++)
        {
            if (!cat[j] && dist[j] < min)
            {
                min = dist[j];
                index = j;
            }
        }
        cat[index] = 1;

        for (int i = 1; i <= n; i++)
        {
            if (!cat[i] && gra[index][i] < INF && dist[i] > dist[index] + gra[index][i])
                dist[i] = dist[index] + gra[index][i];
        }
    }
}
int main()
{
    while (cin >> n >> m)
    {
        memset(cat, 0, sizeof(cat));
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (i == j)
                    gra[i][j] = 0;
                else
                    gra[i][j] = INF;
            }
        }
        int a, b, c;
        for (int i = 0; i < m; i++)
        {
            cin >> a >> b >> c;
            if (gra[a][b] > c) // 少了会wa
                gra[a][b] = gra[b][a] = c;
        }
        if (m == 0)
            cout << "0" << endl;
        else
        {
            Dijkstra(1);
            cout << dist[n] << endl;
        }
    }

    return 0;
}

/***************************************************
User name: jk180602
Result: Accepted
Take time: 44ms
Take Memory: 260KB
Submit time: 2019-01-13 23:32:35
****************************************************/

Floyed：

#include 
#define INF INT_MAX
using namespace std;

int gra[105][105];
int cat[105];
int n, m;

void Floyed()
{
    for (int k = 1; k <= n; k++)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (gra[i][j] > gra[i][k] + gra[k][j])
                    gra[i][j] = gra[i][k] + gra[k][j];
            }
        }
    }
}
int main()
{
    while (cin >> n >> m)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (i == j)
                    gra[i][j] = 0;
                else
                    gra[i][j] = INF;
            }
        }
        int a, b, c;
        while (m--)
        {
            cin >> a >> b >> c;
            if (gra[a][b] > c)
                gra[a][b] = gra[b][a] = c;
        }
        Floyed();
        cout << gra[1][n] << endl;
    }
    return 0;
}

/***************************************************
User name: jk180602
Result: Accepted
Take time: 48ms
Take Memory: 288KB
Submit time: 2019-01-18 11:34:59
****************************************************/