最短路

来源：互联网发布：印度人口知乎编辑：程序博客网时间：2024/05/29 14:32

注：我在修炼当中，博客完全是为了自己复习方便看得，如果你不慎点入了我的博客，看看就好，不要相信，误人子弟就不好了- -

先贴模板

Dijkstra

#define inf 0x3fffffff#define M 105int dist[M], map[M][M], n;bool mark[M];void init (){    int i, j;    for (i = 1; i <= n; i++)    //i==j的时候也可以初始化为0，只是有时候不合适        for (j = 1; j <= n; j++)            map[i][j] = inf;}void dijk (int u){    int i, j, mins, v;    for (i = 1; i <= n; i++)    {        dist[i] = map[u][i];        mark[i] = false;    }    mark[u] = true;    dist[u] = 0;    //既然上面的map当i==j时不是0，就要这句    while (1)    {        mins = inf;        for (j = 1; j <= n; j++)            if (!mark[j] && dist[j] < mins)                mins = dist[j], v = j;        if (mins == inf)            break;        mark[v] = true;        for (j = 1; j <= n; j++)            if (!mark[j] && dist[v] + map[v][j] < dist[j])                dist[j] = dist[v] + map[v][j];    }}

Floyd

#define inf 0x3fffffff  //注意，太大会溢出#define M               //最大点数int n, dist[M][M];           //n：实际点数void init ()            //有时候需要初始化{    int i, j;    for (i = 1; i <= n; i++)        for (j = i + 1; j <= n; j++)            dist[i][j] = dist[j][i] = inf;}void floyd (){    int i, j, k;    for (k = 1; k <= n; k++)        for (i = 1; i <= n; i++)            for (j = 1; j <= n; j++)    //有的题目会溢出就要自己变通了                if (dist[i][k] + dist[k][j] < dist[i][j])                        dist[i][j] = dist[i][k] + dist[k][j];}

SPFA

#define inf 0x3fffffff#define M 1005  //最大点数struct edge{    int v, w, next;} e[10005];     //估计好有多少条边int pre[M], cnt, dist[M], n;bool inq[M];//注意初始化void init (){    cnt = 0;    memset (pre, -1, sizeof(pre));}//注意双向加边void addedge (int u, int v, int w)    //加边函数，慢慢模拟就会明白的{    e[cnt].v = v;    e[cnt].w = w;    e[cnt].next = pre[u];       //接替已有边    pre[u] = cnt++;             //自己前插成为u派生的第一条边}void spfa (int u){    int v, w, i;    for (i = 1; i <= n; i++)    //对于从1到n的编号        dist[i] = inf, inq[i] = false;    dist[u] = 0;    queue<int> q;    q.push (u);    inq[u] = true;    while (!q.empty())    {        u = q.front();        q.pop();        inq[u] = false;        for (i = pre[u]; i != -1; i = e[i].next)        {            w = e[i].w;            v = e[i].v;            if (dist[u] + w < dist[v])            {                dist[v] = dist[u] + w;                if (!inq[v])                {                    q.push (v);                    inq[v] = true;                }            }        }    }}

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