Dijkstra算法——求单源最短路径

来源自我的博客

http://www.yingzinanfei.com/2017/04/07/dijkstrasuanfaqiudanyuanzuiduanlujing/

#include 
#include 
int main(){
    int n, m;
    scanf("%d%d", &n, &m);

    int e[10][10];  // 初始化边
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= m; j++){
            if (i == j) e[i][j] = 0;
            else e[i][j] = INT_MAX;
        }
    }

    int t1, t2, t3; // 输入边
    for (int i = 1; i <= m; i++){
        scanf("%d%d%d", &t1, &t2, &t3);
        e[t1][t2] = t3;
    }

    int dis[10];    // 源点到任意点的距离
    for (int i = 1; i <= n; i++){
        dis[i] = e[1][i];
    }

    int book[10];   // 1表示已处理
    for (int i = 1; i <= n; i++){
        book[i] = 0;
    }
    book[1] = 1;

    // 核心语句，需要处理n-1个点，循环n-1次
    for (int i = 1; i <= n - 1; i++){
        // 找当前未处理点中离源点最近的点
        int min = INT_MAX, u;
        for (int j = 1; j <= n; j++){
            if (book[j] == 0 && dis[j] < min){
                min = dis[j];
                u = j;
            }
        }
        book[u] = 1;    // 将此最近的点标记为已处理

        // 更新此点相邻的点到源点的距离
        for (int v = 1; v <= n; v++){
            if (e[u][v] < INT_MAX){ // 要直接相邻
                if (dis[v] > dis[u] + e[u][v]){
                    dis[v] = dis[u] + e[u][v];
                }
            }
        }
    }

    // 可以得到结果了
    for (int i = 1; i <= n; i++){
        printf("%d ", dis[i]);
    }

    return 0;
}