最短路径

http://acm.hdu.edu.cn/showproblem.php?pid=2544

稍微补充了下 求任意两点间的最短距离

Dijkstra

View Code

 1 #include<stdio.h>

 2 #include<string.h>

 3 #define INF 0x3f3f3f3f

 4 int d[101],w[101][101],f[101];

 5 int Dijkstra(int st,int en,int n)

 6 {

 7     int i,j,min,k;

 8     d[st] = 0;

 9     memset(f,0,sizeof(f));

10     for(i = 1 ;i <= n ; i++)

11         if(i!=st)

12             d[i] = INF;

13     for(i = 1 ;i <= n ; i++)

14     {

15         min = INF;

16         for(j = 1 ;j <= n ;j++)

17             if(!f[j]&&min>=d[j])

18                 min = d[k=j];

19         f[k] = 1;

20         for(j = 1; j <= n ; j++)

21             if(d[j]>d[k]+w[k][j])

22                 d[j] = d[k]+w[k][j];

23     }

24     

25     return d[en];

26 }

27 int main()

28 {

29     int i,j, n, m,x,y,z;

30     while(scanf("%d%d", &n,&m)&&n&&m)

31     {

32         memset(w,INF,sizeof(w));

33         for(i = 1 ;i<= m ;i++)

34         {

35             scanf("%d%d%d", &x,&y,&z);

36             if(w[x][y]>z)

37             {

38                 w[x][y] = z;

39                 w[y][x] = z;

40             }

41         }

42         printf("%d\n",Dijkstra(1,n,n));

43 

44     }

45     return 0;

46 }

Floyd

View Code

 1 #include<stdio.h>

 2 #include<string.h>

 3 #define INF 0x3f3f3f3f

 4 int w[101][101];

 5 int Floyd(int st,int en,int n)

 6 {

 7     int i, j, k;

 8     for(i = 1 ; i <= n ; i++)

 9         w[i][i] = 0;

10     for(i = 1 ;i <= n ; i++)

11         for(j = 1 ; j <= n ; j++)

12             for(k = 1 ; k <= n ; k++)

13                 if(w[j][k]>w[j][i]+w[i][k])

14                     w[j][k] = w[j][i]+w[i][k];

15     return w[st][en];

16 }

17 int main()

18 {

19     int i,j, n, m,x,y,z;

20     while(scanf("%d%d", &n,&m)&&n&&m)

21     {

22         memset(w,INF,sizeof(w));

23         for(i = 1 ;i<= m ;i++)

24         {

25             scanf("%d%d%d", &x,&y,&z);

26             if(w[x][y]>z)

27             {

28                 w[x][y] = z;

29                 w[y][x] = z;

30             }

31         }

32         printf("%d\n",Floyd(1,n,n));

33     }

34     return 0;

35 }

 Bellman－Ford算法

http://www.wutianqi.com/?p=1912

每次松弛把每条边都更新一下，若n-1次松弛后还能更新，则说明图中有负环，无法得出结果，否则就成功完成。Bellman-ford算法有一个小优化：每次松弛先设一个旗帜flag，初值为FALSE，若有边更新则赋值为TRUE，最终如果还是FALSE则直接成功退出。

http://poj.org/problem?id=3259

 

View Code

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<string.h>

 4 #include<queue>

 5 #define INF 0x3f3f3f

 6 using namespace std;

 7 struct node

 8 {

 9     int u,v,t;

10 }q[5000];

11 int dis[501];

12 int bellford(int n,int m)

13 {

14     int i,j;

15     memset(dis,0,sizeof(0));

16     for(i = 1; i <= n ; i++)

17     {

18         int flag = 0;

19         for(j = 1 ; j <= m ; j++)

20         if(dis[q[j].v]>dis[q[j].u]+q[j].t)

21         {

22             dis[q[j].v] = dis[q[j].u]+q[j].t;

23             flag = 1;

24         }

25         if(!flag)

26         break;

27     }

28     for(j = 1; j <= m ; j++)

29     {

30         if(dis[q[j].v]>dis[q[j].u]+q[j].t)

31             return 1;

32     }

33     return 0;

34 }

35 int main()

36 {

37     int i,j,k,n,m,t,w1,o,a,b,c;

38     scanf("%d",&t);

39     while(t--)

40     {

41         scanf("%d%d%d",&n,&m,&w1);

42         o = 0;

43         for(i = 1 ; i <= m ; i++)

44         {

45             scanf("%d%d%d",&a,&b,&c);

46             o++;

47             q[o].u = a;

48             q[o].v = b;

49             q[o].t = c;

50             o++;

51             q[o].u = b;

52             q[o].v = a;

53             q[o].t = c;

54         }

55         for(i = 1 ; i <= w1 ; i++)

56         {

57             scanf("%d%d%d",&a,&b,&c);

58             o++;

59             q[o].u = a;

60             q[o].v = b;

61             q[o].t = -c;

62         }

63        if(bellford(n,o))

64        printf("YES\n");

65        else

66        printf("NO\n");

67     }

68     return 0;

69 }