WA了好多次... 这题要用long long 而且INF要设大一点
Sample Input
2 //T
1 2 3 4 1 3 5 7 //L1-L4 C1-C4 距离和花费
4 2 //结点数 询问次数
1 //结点的横坐标
2
3
4
1 4 //起点 终点
4 1
1 2 3 4 1 3 5 7
4 1
1
2
3
10
1 4
Sample Output
Case 1:
The minimum cost between station 1 and station 4 is 3.
The minimum cost between station 4 and station 1 is 3.
Case 2:
Station 1 and station 4 are not attainable.
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 # include <cmath> 6 # include <queue> 7 # define LL long long 8 using namespace std ; 9 10 const LL INF=0x7f7f7f7f7f7f7f7fLL; 11 const int MAXN=210; 12 13 LL L[10] ; 14 LL C[10] ; 15 LL x[MAXN] ; 16 17 LL dis[MAXN][MAXN]; 18 int n ; 19 20 21 void floyed()//节点从1~n编号 22 { 23 int i,j,k; 24 for(k=1;k<=n;k++) 25 for(i=1;i<=n;i++) 26 for(j=1;j<=n;j++) 27 if(dis[i][k]+dis[k][j] < dis[i][j] && dis[i][k] != INF && dis[k][j] != INF) 28 dis[i][j]=dis[i][k]+dis[k][j]; 29 30 } 31 32 LL Cost(LL d) 33 { 34 if (d < 0) 35 d *= -1 ; 36 if (d > 0 && d<= L[1]) 37 return C[1] ; 38 if (d > L[1] && d<= L[2]) 39 return C[2] ; 40 if (d > L[2] && d<= L[3]) 41 return C[3] ; 42 if (d > L[3] && d<= L[4]) 43 return C[4] ; 44 return INF ; 45 } 46 47 int main () 48 { 49 // freopen("in.txt","r",stdin) ; 50 int cnt ; 51 int T ; 52 cin>>T ; 53 int Case = 0 ; 54 while (T--) 55 { 56 Case++ ; 57 cout<<"Case "<<Case<<":"<<endl ; 58 int i , j ; 59 LL w ; 60 for (i = 1 ; i <= 4 ; i++) 61 cin>>L[i]; 62 for (i = 1 ; i <= 4 ; i++) 63 cin>>C[i]; 64 65 cin>>n>>cnt ; 66 for (i = 1 ; i <= n ; i++) 67 cin>>x[i]; 68 for (i = 1 ; i <= n ; i++) 69 for (j = 1 ; j <= n ; j++) 70 { 71 if(i==j)dis[i][j]=0; 72 else dis[i][j]=INF; 73 } 74 for (i = 1 ; i <= n ; i++) 75 for (j = i+1 ; j <= n ; j++) 76 { 77 LL d = x[i] - x[j] ; 78 w = Cost(d) ; 79 dis[i][j] = w ; 80 dis[j][i] = w ; 81 } 82 floyed() ; 83 int u , v ; 84 while(cnt--) 85 { 86 cin>>u>>v ; 87 if (dis[u][v] != INF) 88 cout<<"The minimum cost between station "<<u<<" and station "<<v<<" is "<<dis[u][v]<<"."<<endl ; 89 else 90 cout<<"Station "<<u<<" and station "<<v<<" are not attainable."<<endl ; 91 } 92 } 93 94 return 0 ; 95 }