A,C,D,E都是Floyd算法........代码就不发了。
代码： 328MS
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 using namespace std; 5 const int MAX = 1000+5,inf = 1061109568; 6 int map[MAX][MAX][2],visit[MAX],T[MAX],cost[MAX],n,m; 7 int main() 8 { 9 int a,b,d,p,s,e; 10 while(scanf("%d%d",&n,&m),n+m) 11 { 12 memset(map,0x3f,sizeof(map)); 13 memset(cost,0x3f,sizeof(cost)); 14 memset(T,0x3f,sizeof(T)); 15 for(int i = 1;i<=m;++i) 16 { 17 scanf("%d%d%d%d",&a,&b,&d,&p); 18 if(d<map[a][b][0]) 19 { //对于输入的处理很重要 20 map[a][b][0] = map[b][a][0] = d; 21 map[a][b][1] = map[b][a][1] = p; 22 } 23 else if(d==map[a][b][0]) 24 map[a][b][1] = map[b][a][1]=min(map[a][b][1],p); 25 } 26 scanf("%d%d",&s,&e); 27 T[s] = cost[s] = 0; //将起点的花费和路径长度都赋0 28 T[0] = inf-1; //为便于更新s的值，将T[0]赋无穷大 29 memset(visit,0,sizeof(visit)); 30 while(1) 31 { 32 for(int i = 1;i<=n;++i) 33 if(!visit[i]&&(T[s]+map[s][i][0])<=T[i]) 34 { 35 if(T[s]+map[s][i][0]<T[i]) 36 { //这一步的处理很重要，要将等于和小于的情况分开处理 37 T[i] = T[s] + map[s][i][0]; 38 cost[i] = cost[s] + map[s][i][1]; 39 } 40 else if(T[s]+map[s][i][0]==T[i]) 41 cost[i] = min(cost[i],cost[s] + map[s][i][1]); 42 } 43 visit[s] = 1; 44 s = 0; //将s做初始化处理 45 bool flag = 1; 46 for(int i = 1;i<=n;++i) 47 if(!visit[i]&&T[i]<T[s]) 48 { 49 flag = 0; 50 s = i; 51 } 52 if(flag) //设置退出条件 53 break; 54 } 55 printf("%d %d ",T[e],cost[e]); 56 } 57 return 0; 58 }
H 凶残的看不懂题的题......还是Floyd Tram POJ 1847 16ms
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 using namespace std; 6 const int maxn = 110; 7 const int INF = 10000000; 8 int dist[maxn][maxn]; 9 int n,start, end; 10 int main() 11 { 12 while(~scanf("%d%d%d",&n,&start,&end)) 13 { 14 for(int i=1;i<=n;i++) 15 { 16 for(int j=1;j<=n;j++) 17 if(i==j) 18 dist[i][j]=0; 19 else 20 dist[i][j] = INF; 21 } 22 int path,y; 23 for(int i=1;i<=n;i++) 24 { 25 scanf("%d",&path); 26 for(int j=1;j<=path;j++) 27 { 28 scanf("%d",&y); 29 if(j==1) 30 dist[i][y]=0; 31 else 32 dist[i][y]=1; 33 } 34 } 35 for(int i=1;i<=n;i++) 36 for(int j=1;j<=n;j++) 37 for(int k=1;k<=n;k++) 38 dist[j][k] =min(dist[j][k],dist[j][i] + dist[i][k] ); 39 } 40 if(dist[start][end]==INF) 41 printf("-1 "); 42 else 43 printf("%d ",dist[start][end]); 44 return 0; 45 }
代码： 0ms
1 #include <stdio.h> 2 #include <string.h> 3 const int INF = 0x3f3f3f3f; 4 int map[110][110]; 5 int rank[110],dis[110],vis[110]; 6 int n,m; 7 int Dijkstra(int k) 8 { 9 int i,j; 10 memset(vis,0,sizeof(vis)); 11 vis[0] = 1; 12 for(i = 0; i <= n; i++) 13 dis[i] = map[0][i]; 14 dis[0] = 0; 15 for(i = 1; i <= n; i++) 16 { 17 int min = INF,pos; 18 for(j = 0; j <= n; j++) 19 if(dis[j] < min && !vis[j]) 20 min = dis[pos=j]; 21 vis[pos] = 1; 22 for(j = 0; j <= n; j++) 23 if(!vis[j] && dis[j] > dis[pos] + map[pos][j] && rank[pos] >= k && rank[pos] <= k+m) 24 dis[j] = dis[pos] + map[pos][j]; 25 } 26 return dis[1]; 27 } 28 int main() 29 { 30 int i; 31 int p,x,t,v; 32 memset(map, 0x3f, sizeof(map)); 33 scanf("%d %d",&m,&n); 34 for(i = 1; i <= n; i++) 35 { 36 scanf("%d %d %d",&p,&rank[i],&x); 37 map[0][i] = p; 38 while(x--) 39 { 40 scanf("%d %d",&t,&v); 41 map[t][i] = v; //能替代i的物品，编号为t，价值为v 42 } 43 } 44 int min = INF,ans; 45 for(i = rank[1]-m; i <= rank[1]; i++) //枚举从rank[1]-m到rank[1]的等级；取这些最短路中最短的； 46 { 47 ans = Dijkstra(i); 48 if(min > ans) 49 min = ans; 50 } 51 printf("%d ",min); 52 return 0; 53 }