月球美容計劃之最短路

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那HDU的2544作為複習最短路的題目，用不同演算法。

迪傑斯特拉

有點像普利姆演算法的精簡版，不能有負權邊

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX 99999#define qmin(a,b) a > b ? b : a//最短路//迪傑斯特拉int G[200][200];int vis[200];int djs (int n,int s){    int d[200];    memset(vis,0,sizeof(vis));    int i,k;    for (i = 1;i <= n;i++)        d[i] = G[1][i];    d[1] = 0;    vis[1] = 1;    int imin,xb = 1;    for (i = 1;i < n;i++)    {        imin = MAX;        for (k = 1;k <= n;k++)//以xb為起點↓            if (!vis[k] && d[xb] + G[xb][k] < d[k])  //最短的那條邊，快到碗裡來                d[k] = d[xb] + G[xb][k];        vis[xb] = 1;        for (k = 1;k <= n;k++)            if (!vis[k] && imin > d[k])                imin = d[xb = k];           //找到最小的點，並以此為起點找最短        vis[xb] = 1;    }    return d[n];}int main(){    int n,m;    while (scanf ("%d%d",&n,&m),n || m)    {        int i,k;        for (i = 0;i <= n;i++)            for (k = 0;k <= n;k++)                if (i == k)                    G[i][k] = 0;                else                    G[i][k] = MAX;        for (i = 0;i < m;i++)        {            int a,b,c;            scanf ("%d%d%d",&a,&b,&c);            if (c < G[a][b])            {                G[a][b] = c;                G[b][a] = c;            }        }        int ans = djs(n,1);        printf ("%d\n",ans);    }    return 0;}

 

貝爾曼福特

能夠有負權邊，就是不停的鬆弛，時間複雜度有點高

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX 99999#define qmin(a,b) a > b ? b : a//最短路//貝爾曼福特struct E{    int e,v;    int w;}e[10000];int cont;int BF (int n,int s){    int d[200];    int i,k;    for (i = 0;i <= n;i++)        d[i] = MAX;    d[s] = 0;    for (i = 1;i < n;i++)     //找n - 1條邊    {        for (k = 0;k < cont;k++)  //把每條邊都遍曆一遍        {            int a = e[k].e,b = e[k].v;            d[b] = qmin (d[b],d[a] + e[k].w);  //鬆弛        }    }    return d[n];}int main(){    int n,m;    while (scanf ("%d%d",&n,&m),n || m)    {        int i,k;        cont = 0;        for (i = 0;i < m;i++)        {            int a,b,c;            int tf = 1;            scanf ("%d%d%d",&a,&b,&c);            for (k = 0;k < cont;k++)                if ((e[k].e == a && e[k].v == b) || (e[k].e == b && e[k].v == a))                    if (e[k].w > c)                    {                        e[k].w = c;                        tf = 0;                        break;                    }            if (tf)            {//無向圖                e[cont].e = a;                e[cont].v = b;                e[cont++].w = c;                e[cont].e = b;                e[cont].v = a;                e[cont++].w=c;            }        }        int ans = BF(n,1);        printf ("%d\n",ans);    }    return 0;}

 

弗洛伊德

這簡直就像暴力啊，時間複雜度大的誇張，可是求出了每兩個點之間的最短路，對於資料多並且圖小的題目還是能夠考慮的。

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX 99999#define qmin(a,b) a > b ? b : a//最短路//弗洛伊德int G[200][200];int fld (int n,int s){    int d[200][200];    int i,j,k;    memcpy(d,G,sizeof (G));    for (i = 0;i <= n;i++)        for (k = 0;k <= n;k++)            for (j = 0;j <= n;j++)                d[k][j] = qmin (d[k][j],d[k][i] + d[i][j]);    return d[s][n];}int main(){    int n,m;    while (scanf ("%d%d",&n,&m),n || m)    {        int i,k;        //鄰接矩陣初始化        for (i = 0;i <= n;i++)            for (k = 0;k <= n;k++)                if (i == k)                    G[i][k] = 0;                else                    G[i][k] = MAX;        for (i = 0;i < m;i++)        {            int a,b,c;            scanf ("%d%d%d",&a,&b,&c);            if (G[a][b] > c)            {                G[a][b] = c;                G[b][a] = c;            }        }        int ans = fld(n,1);        printf ("%d\n",ans);    }    return 0;}

 

SPFA

BF的隊列最佳化，有點像BFS。

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX 99999#define qmin(a,b) a > b ? b : a//最短路//SPFAstruct node{    int v;    int w;    struct node *next;}head[10001];int q[1000000];   //隊列int s = 0,e = 0;int SPFA (int n,int st){    bool inq[200];  //標記是否還在隊列中（隊列中的不在入隊列）    int d[200];    int i;    memset (inq,0,sizeof (inq));    for (i = 0;i <= n;i++)        d[i] = MAX;    s = 0;    e = 0;    d[st] = 0;    q[s++] = st;    inq[st] = 1;    //源點入隊列並標記    while (s > e)    {        int now = q[e++];        inq[now] = 0;  //出隊列的就恢複標記        struct node *p = head[now].next;        while (p != NULL)        {            if(d[p->v] > d[now] + p->w)  //鬆弛成功            {                d[p->v] = d[now] + p->w;                if (!inq[p->v])         //假設在隊列中就不入隊列                {                    q[s++] = p->v;                    inq[p->v] = 1;                }            }            p = p->next;        }    }    return d[n];          //其它的和BF一樣了}int add (int a,int b,int c)    //鄰接鏈表{    struct node *t = new node;    t->v = b;    t->w = c;    t->next = NULL;    struct node *p = &head[a];    while (p->next != NULL)        p = p->next;    p->next = t;    return 1;}int main(){    int n,m;    while (scanf ("%d%d",&n,&m),n || m)    {        memset(head,0,sizeof (head));        int i,k;        for (i = 0;i < m;i++)        {            int a,b,c;            int tf = 1;            scanf ("%d%d%d",&a,&b,&c);            add (a,b,c);            add (b,a,c);        }        int ans = SPFA(n,1);        printf ("%d\n",ans);    }    return 0;}

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