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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3031 Accepted Submission(s):
940
e (German for a one-way street) is
a street on which vehicles should only move in one direction. One reason for
having one-way streets is to facilitate a smoother flow of traffic through
crowded areas. This is useful in city centers, especially old cities like Cairo
and Damascus. Careful planning guarantees that you can get to any location
starting from any point. Nevertheless, drivers must carefully plan their route
in order to avoid prolonging their trip due to one-way streets. Experienced
drivers know that there are multiple paths to travel between any two locations.
Not only that, there might be multiple roads between the same two locations.
Knowing the shortest way between any two locations is a must! This is even more
important when driving vehicles that are hard to maneuver (garbage trucks,
towing trucks, etc.)
n 个地点（包括公司的车库），c 表示c辆车抛锚的地点， r 条道路
第二行给出 c+1 个地点，第一个为车库地点， 其余的 c 个为车的地点。
接下来的 r 行表示 r 条有向的道路，
s1 -- d -> s2 表示 s1到s2 的长度为 d
s1 <- d -- s2 表示 s2到s1 的长度为 d
s1 <- d -> s2 表示 s1到s2为双向边， 且长度为 d
拖车从车库出发到每个地点，在该地点拖回抛锚的车子。一辆拖车一次只能拖回一辆车子。注意：同一个地点可能有多辆抛锚的车子求总的路径长度。还要输出 case number
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <map> 5 #define inf 0x3f3f3f3f 6 #define M 105 7 using namespace std; 8 int n,c,m,t; 9 int maps[M][M]; 10 char str[1005][M]; 11 map<string,int> mat; 12 13 void floyd() //小于100，使用floyd处理 14 { 15 int i,j,k; 16 for(k=1; k<=n; k++) 17 for(i=1; i<=n; i++) 18 for(j=1; j<=n; j++) 19 if(maps[i][j]>maps[i][k]+maps[k][j]) 20 maps[i][j]=maps[i][k]+maps[k][j]; 21 } 22 23 int main() 24 { 25 int i,j,s,x,y,start,Case=1; 26 char s1[M],s2[M]; 27 char from,to; 28 while(~scanf("%d%d%d",&n,&c,&m)) 29 { 30 if(!n&&!m&&!c) 31 break; 32 int sum=0; 33 mat.clear(); //map函数清空 34 for(i=0; i<=n; i++) 35 for(j=0; j<=n; j++) 36 if(i==j) maps[i][j]=0; 37 else maps[i][j]=inf; 38 for(i=0; i<=c; i++) 39 scanf("%s",str[i]); 40 t=1; 41 for(i=0; i<m; i++) 42 { 43 scanf("%s %c-%d-%c %s",s1,&from,&s,&to,s2); //注意这里的输入 44 if(!mat[s1]) mat[s1]=t++; //运用了map函数，假如s1不在mat中，则存入，并加1 45 if(!mat[s2]) mat[s2]=t++; 46 x=mat[s1],y=mat[s2]; 47 if(from==‘<‘&&s<maps[y][x]) 48 maps[y][x]=s; 49 if(to==‘>‘&&s<maps[x][y]) 50 maps[x][y]=s; 51 } 52 floyd(); 53 start=mat[str[0]]; //找寻起点在mat中存储的位置 54 for(i=1; i<=c; i++) 55 sum+=maps[start][mat[str[i]]]+maps[mat[str[i]]][start]; //注意去和回！！ 56 printf("%d. %d\n",Case++,sum); 57 } 58 return 0; 59 }
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原文地址：http://www.cnblogs.com/pshw/p/5435645.html
