Einbahnstrasse

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3469    Accepted Submission(s): 1096 点击打开题目链接 Problem Description Einbahnstra   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.) You just started a new job at a car-towing company. The company has a number of towing trucks parked at the company's garage. A tow-truck lifts the front or back wheels of a broken car in order to pull it straight back to the company's garage. You receive calls from various parts of the city about broken cars that need to be towed. The cars have to be towed in the same order as you receive the calls. Your job is to advise the tow-truck drivers regarding the shortest way in order to collect all broken cars back in to the company's garage. At the end of the day, you have to report to the management the total distance traveled by the trucks.   Input Your program will be tested on one or more test cases. The first line of each test case specifies three numbers (N , C , and R ) separated by one or more spaces. The city has N locations with distinct names, including the company's garage. C is the number of broken cars. R is the number of roads in the city. Note that 0 < N < 100 , 0<=C < 1000 , and R < 10000 . The second line is made of C + 1 words, the first being the location of the company's garage, and the rest being the locations of the broken cars. A location is a word made of 10 letters or less. Letter case is significant. After the second line, there will be exactly R lines, each describing a road. A road is described using one of these three formats: A -v -> B A <-v - B A <-v -> B A and B are names of two different locations, while v is a positive integer (not exceeding 1000) denoting the length of the road. The first format specifies a one-way street from location A to B , the second specifies a one-way street from B to A , while the last specifies a two-way street between them. A , ``the arrow", and B are separated by one or more spaces. The end of the test cases is specified with a line having three zeros (for N , C , and R .) The test case in the example below is the same as the one in the figure.    Output For each test case, print the total distance traveled using the following format: k . V Where k is test case number (starting at 1,) is a space, and V is the result.   Sample Input 4 2 5 NewTroy Midvale Metrodale NewTroy <-20-> Midvale Midvale --50-> Bakerline NewTroy <-5-- Bakerline Metrodale <-30-> NewTroy Metrodale --5-> Bakerline 0 0 0   Sample Output 1. 80

题目意思：输入数据第一行给出N，代表N个地点，C,C个车坏掉的地点，R,R条直通道路。 接下来给出C+1个字符串，其中第一个代表汽车修理总部，剩余C个代表车发生故障的地点。 接下来R行，给出R条直通道路，形式是 （地点1   关系   地点2），关系包含两者的互通 关系和距离。关系中仅有'>'代表这条路是从地点1到地点2的单向道路，关系中仅存在'<'代表这条路是从地点2到地点1的单向道路。 关系中既有'>'又有'<'代表这条路是双向道路，可以从地点1到地点2，也可以从地点2到地点1. 求解从汽车修理总部将所有坏掉的车运回来的最小花费。 与此题相似的题目有：POJ 1511题目 Invitation Cards（不过这个题目节点数目比较多，难度较这个题目大） 解题思路： 难点：将字符串对应与整形的节点编号，使用stl的map映射。同类型的题目还有杭电的2112题HDU Today.也是牵涉到要将字符串 表示的节点转化为整型编号。 小提示：题目中已指出，两个点之间可能存在多条道路，所以一定要考虑取最短的那一条路；很多人会想如果多辆车在同一地点坏掉， 可以同时将这些车拖回来，经过亲身实验证明，这样的做法提交后结果为Wrong Answer，说明每次只能拖一辆坏的车回去；一定要 注意这个题目的图是有向图。如果从汽车修理部到某个点运汽车，则去的路和回来的路不一定是相同的。mincost = 洗车修理部 到每个车坏的点的距离 + 每个车坏掉的点到汽车修理部的距离.在此提供两种解题方法。 方法1：构建两个图，一个图是题中给出的原图，第二个图为与题中给出图完全相反的图。 因为输入的时候第一个字符串是汽车修理总部，则按照代码，汽车修理总部的节点编号为1，则先在原图上求出节点1到其他节点的最短 路径。然后再在反向图中求1到其他点的最短路径。（相当于求了其他点到1的最短路径，仔细想想是不是这样）

#include<iostream> #include<cstdio> #include<queue> #include<vector> #include<cstring> #include<string> #include<stack> #include<map> #include<set> #define INF 0x3f3f3f3f #define MAXN 102 using namespace std; int Map[MAXN][MAXN][2]; ///Map[MAXN][MAXN][0]用来存放正向图，Map[MAXN][MAXN][1]用来存放反向图 int dis[MAXN][2],vis[MAXN]; ///dis[MAXN][0]用来存放正向图，节点1到各个点的最短路径。 ///dis[MAXN][1]用来存放反向图，求节点1到各个点的最短路径。相当于求各个点到1的最短路径。 ///vis是用来标记的数组。 int N,C,R; void InitMap() ///初始化地图 { for(int k = 0; k < 2; k++) { for(int i = 1; i <= N; i++) { Map[i][i][k] = 0; for(int j = i+1; j <= N; j++) Map[i][j][k] = Map[j][i][k] = INF; } } } void Dijkstra(int a) ///参数a代表求那个图的最短路径，0代表求原图，1代表求反图 { int mindis,u; for(int i = 1; i <= N; i++) { vis[i] = 0; dis[i][a] = Map[1][i][a]; } vis[1] = 1; for(int i = 1; i <= N; i++) { mindis = INF; u = 0; for(int j = 1; j <= N; j++) { if(vis[j]==0 && dis[j][a]<mindis) { u = j; mindis = dis[j][a]; } } if(u == 0) break; vis[u] = 1; for(int j = 1; j <= N; j++) { if(vis[j]==0) { if(dis[u][a]+Map[u][j][a]<dis[j][a]) { dis[j][a] = dis[u][a]+Map[u][j][a]; } } } } } int main() { int t = 0; while(~scanf("%d%d%d",&N,&C,&R)) ///N是location的个数，C是出故障的车的数目，R是直通道路条数 { if(N==0 && C==0 && R==0) break; ///程序结束条件 InitMap(); ///完成图的初始化工作 map<string,int>m; vector<int>v; char pos1[12],pos2[12],dist[50]; int num = 0; for(int i = 0; i <= C; i++) { scanf("%s",pos1); ///输入城市。 if(!m[pos1]) { m[pos1] = (++num); } v.push_back(m[pos1]); } int _go,_back,sub1,sub2,_dist; for(int i = 0; i < R; i++) { _go = _back = _dist = 0; scanf("%s%s%s",pos1,dist,pos2); if(!m[pos1]) m[pos1] = (++num); sub1 = m[pos1]; if(!m[pos2]) m[pos2] = (++num); sub2 = m[pos2]; for(int j = 0; j < strlen(dist); j++) { if(dist[j]=='>') _go = 1; if(dist[j]=='<') _back = 1; if(dist[j]>='0'&&dist[j]<='9') _dist = _dist*10 + (dist[j]-'0'); } if(_go) ///sub1到sub2有路 { if(_dist < Map[sub1][sub2][0]) { Map[sub1][sub2][0] = _dist; Map[sub2][sub1][1] = _dist; } } if(_back) ///sub2到sub1有路 { if(_dist < Map[sub2][sub1][0]) { Map[sub2][sub1][0] = _dist; Map[sub1][sub2][1] = _dist; } } } Dijkstra(0); Dijkstra(1); int ans = 0; for(int i = 0; i < v.size(); i++) { ans = ans + dis[v[i]][0] + dis[v[i]][1]; } printf("%d. %d\n",++t,ans); } return 0; } 方法2：直接采用Flody算法来计算任意两点间的最短路径。不过一开始害怕超时，放弃了，后来试了试，也是可以过滴。

#include<iostream> #include<cstdio> #include<queue> #include<vector> #include<cstring> #include<string> #include<stack> #include<map> #include<set> #define INF 0x3f3f3f3f #define MAXN 102 using namespace std; int Map[MAXN][MAXN]; int N,C,R; void InitMap() { for(int i = 1; i <= N; i++) { Map[i][i] = 0; for(int j = i+1; j <= N; j++) { Map[i][j] = Map[j][i] = INF; } } } void Flody() { for(int k = 1; k <= N; k++) { for(int i = 1; i <= N; i++) { for(int j = 1; j <= N; j++) { if(Map[i][k]+Map[k][j] < Map[i][j]) Map[i][j] = Map[i][k]+Map[k][j]; } } } } int main() { int t = 0; while(~scanf("%d%d%d",&N,&C,&R)) ///N是location的个数，C是出故障的车的数目，R是直通道路条数 { if(N==0 && C==0 && R==0) break; ///程序结束条件 InitMap(); ///完成图的初始化工作 map<string,int>m; vector<int>v; ///用容器保存汽车出故障的地点 char pos1[12],pos2[12],dist[50]; int num = 0; for(int i = 0; i <= C; i++) { scanf("%s",pos1); ///输入城市。 if(!m[pos1]) { m[pos1] = (++num); } v.push_back(m[pos1]); } int _go,_back,sub1,sub2,_dist; for(int i = 0; i < R; i++) { _go = _back = _dist = 0; scanf("%s%s%s",pos1,dist,pos2); if(!m[pos1]) m[pos1] = (++num); sub1 = m[pos1]; if(!m[pos2]) m[pos2] = (++num); sub2 = m[pos2]; for(int j = 0; j < strlen(dist); j++) { if(dist[j]=='>') _go = 1; if(dist[j]=='<') _back = 1; if(dist[j]>='0'&&dist[j]<='9') _dist = _dist*10 + (dist[j]-'0'); } if(_go) ///sub1到sub2有路 { if(_dist < Map[sub1][sub2]) { Map[sub1][sub2] = _dist; } } if(_back) ///sub2到sub1有路 { if(_dist < Map[sub2][sub1]) { Map[sub2][sub1] = _dist; } } } Flody(); int ans = 0; for(int i = 0; i < v.size(); i++) { ans = ans + Map[1][v[i]] + Map[v[i]][1]; } printf("%d. %d\n",++t,ans); } return 0; }