05-图3. 六度空间 (30) floyd bfs

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05-图3. 六度空间 (30) 题目地址

floyd 最后一个case 超时（想想也应该是超时的）

/*05-图3. 六度空间 (30)http://www.patest.cn/contests/mooc-ds/05-%E5%9B%BE3floyd() 算所有两点之间的距离 运行超时*/#include <iostream>  #define N 10001#define INF 9999999int mp[N][N];int num[N];int n, m;void floyd(){    int i, j, k;    for (i = 1; i <= n; i++)    {        for (j = 1; j <= n; j++)        {            for (k = 1; k <= n; k++)            {                if (mp[i][k] + mp[k][j] < mp[i][j])                {                    mp[i][j] = mp[i][k] + mp[k][j];                }            }            if (mp[i][j] <= 6)            {                num[i] ++;            }        }    }    /*for (i = 1; i <= n; i++)    {    for (j = 1; j <= n; j++)    {    if (mp[i][j] <= 6)    }    }*/}int main(){    //freopen("in", "r", stdin);    while (scanf("%d%d", &n, &m) != EOF)    {        int i, j;        for (i = 1; i <= n; i++)        {            for (j = 1; j <= n; j++)            {                mp[i][j] = INF;            }            mp[i][i] = 0;            num[i] = 0;        }        int u, v;        for (i = 0; i < m; i++)        {            scanf("%d%d", &u, &v);            mp[u][v] = 1;            mp[v][u] = 1;        }        floyd();        for (i = 1; i <= n; i++)        {            printf("%d: ", i);            double sum = num[i] * 1.0 / n;            printf("%.2f%%\n", sum*100);        }        //printf("\n");    }    return 0;}

采用bfs 层次遍历可以 ac

/*05-图3. 六度空间 (30)http://www.patest.cn/contests/mooc-ds/05-%E5%9B%BE3bfs()层次遍历  利用queue 和 lastVal 变量来标记每一层直接ac*/#include <iostream>  #include <queue>using namespace std;#define N 10001int n, m;vector<int> mp[N]; // mp[i] i 可到达的点的vector<> 遍历vector 就可以取得相应的点bool visit[N];int bfs(int sta){    int i, j, k;    for (i = 1; i <= n; i++)    {        visit[i] = false; // visit 初始化    }    queue<int> que;    que.push(sta);    visit[sta] = true;    int lastVal = sta;    int num = 0;    int count = 0;    while (num <= 6 && !que.empty())    {        vector<int> v; // 用来得到当前层数的所有元素        v.clear();        int tmp = que.front();        while (tmp != lastVal)        {            v.push_back(tmp);             que.pop();            if (que.empty())            {                break;            }            else{                tmp = que.front();            }        }        if (tmp == lastVal)        {            v.push_back(tmp);            que.pop();        }        int len = v.size(); // 当前层数的元素个数          if (len <= 0) // 若为0 说明 已经不需要查找下去 直接退出        {            break;        }        count += len; // 记录数加上1         num++; // 层数加上 1        for (i = 0; i < len; i++)        {            int nowVal = v[i];            int nowValLinkLen = mp[nowVal].size();            for (j = 0; j < nowValLinkLen; j++)            {                if (!visit[mp[nowVal][j]])                {                    que.push(mp[nowVal][j]);                    lastVal = mp[nowVal][j];                    visit[mp[nowVal][j]] = true;                }            }        }    }    return count;}int main(){    //freopen("in", "r", stdin);    while (scanf("%d%d", &n, &m) != EOF)    {        int i, j;        for (i = 1; i <= n; i++)        {            mp[i].clear();        }        int u, v;        for (i = 0; i < m; i++)        {            scanf("%d%d", &u, &v);            mp[u].push_back(v);            mp[v].push_back(u);        }        for (i = 1; i <= n; i++)        {            printf("%d: ", i);            double sum = bfs(i) * 1.0 / n;            printf("%.2f%%\n", sum*100);        }        //printf("\n");    }    return 0;}

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