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
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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
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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 }