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  • HDU 2544 最短路 http://acm.hdu.edu.cn/showproblem.php?pid=2544

    //代码:
    
    //方法1:Dijkstra's Algorithm
    
    #include<stdio.h>
    #include<math.h>
    #include<string.h>
    #define INF 0xfffffff
    #define N 110
    #define min(a, b)(a < b ? a : b)
    
    int maps[N][N];
    int d[N], visit[N];
    int n, m;
    
    void Init()
    {
        int i, j;
        memset(visit, 0, sizeof(visit));
        for(i = 1 ; i <= n ; i++)
        {
            d[i] = INF;
            for(j = 1 ; j <= n ; j++)
                maps[i][j] = INF;
        }
    }
    
    void Dij()
    {
        int i, j, x, min;
        d[1] = 0;
        for(i = 1 ; i <= n ; i++)
        {
            min = INF;
            for(j = 1 ; j <= n ; j++)
            {
                if(!visit[j] && d[j] < min)
                {
                    min = d[j];
                    x = j;
                }
            }
            visit[x] = 1;
            for(j = 1 ; j <= n ; j++)
            {
                if(!visit[j] && d[j] > d[x] + maps[x][j])
                    d[j] = d[x] + maps[x][j];
            }
        }
    }
    
    int main()
    {
        int i, a, b, c;
        while(scanf("%d%d", &n, &m), m + n != 0)
        {
            Init();
            for(i = 1 ; i <= m ; i++)
            {
                scanf("%d%d%d", &a, &b, &c);
                maps[a][b] = min(maps[a][b], c);
                maps[b][a] = maps[a][b];
            }
            Dij();
            printf("%d
    ", d[n]);
        }
        return 0;
    }//单元最短路
    
    //方法2:弗洛伊德算法
    
    #include<stdio.h>
    #include<string.h>
    #define N 110
    #define INF 0xfffffff
    #define min(a, b) (a < b ? a : b)
    
    int maps[N][N];
    int n, m;
    
    void Init()
    {
        int i, j;
        for(i = 1 ; i <= n ; i++)
            for(j = 1 ; j <= n ; j++)
                maps[i][j] = INF;
    }
    
    void Floyd()
    {
        int i, k, j;
        for(k = 1 ; k <= n ; k++)
        {
            for(i = 1 ; i <= n ; i++)
            {
                for(j = 1 ; j <= n ; j++)
                    maps[i][j] = min(maps[i][j], maps[i][k] + maps[k][j]);
            }
        }
    }
    
    int main()
    {
        int a, b, c, i;
        while(scanf("%d%d", &n, &m), m + n != 0)
        {
            Init();
            for(i = 1 ; i <= m ; i++)
            {
                scanf("%d%d%d", &a, &b, &c);
                maps[a][b] = min(maps[a][b], c);
                maps[b][a] = maps[a][b];
            }
            Floyd();
            printf("%d
    ", maps[1][n]);
        }
        return 0;
    }//多元最短路
    View Code
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  • 原文地址:https://www.cnblogs.com/qq2424260747/p/4437769.html
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