1003 Emergency (25分) && L2-001 紧急救援 (25分) C++ 邻接表实现

Posted on 2020-09-10 17:28  AlexMaxes  阅读(138)  评论(0)    收藏  举报

两个题目是一样的 中文的多了一点点小需求

L2-001 紧急救援 (25分)
 

作为一个城市的应急救援队伍的负责人，你有一张特殊的全国地图。在地图上显示有多个分散的城市和一些连接城市的快速道路。每个城市的救援队数量和每一条连接两个城市的快速道路长度都标在地图上。当其他城市有紧急求助电话给你的时候，你的任务是带领你的救援队尽快赶往事发地，同时，一路上召集尽可能多的救援队。

输入格式:

输入第一行给出4个正整数N、M、S、D，其中N（2）是城市的个数，顺便假设城市的编号为0 ~ (；M是快速道路的条数；S是出发地的城市编号；D是目的地的城市编号。

第二行给出N个正整数，其中第i个数是第i个城市的救援队的数目，数字间以空格分隔。随后的M行中，每行给出一条快速道路的信息，分别是：城市1、城市2、快速道路的长度，中间用空格分开，数字均为整数且不超过500。输入保证救援可行且最优解唯一。

输出格式:

第一行输出最短路径的条数和能够召集的最多的救援队数量。第二行输出从S到D的路径中经过的城市编号。数字间以空格分隔，输出结尾不能有多余空格。

输入样例:

4 5 0 3
20 30 40 10
0 1 1
1 3 2
0 3 3
0 2 2
2 3 2

输出样例:

2 60
0 1 3

MYCODE:
Dijstra算法，为了实现多条最短路径进行了一点修改




  1 #include <iostream>
  2 #include <vector>
  3 #include <limits.h> //INT_MAX
  4 using namespace std;
  5 
  6 #define MaxSize 500
  7 
  8 typedef struct Road
  9 {
 10     int distance;
 11     int City;
 12     Road *nextRoad;
 13     /* data */
 14 } Road, *ptrRoad;
 15 
 16 typedef struct
 17 {
 18     int man;
 19     ptrRoad firstRoad;
 20 } City, *ptrCity;
 21 
 22 typedef struct
 23 {
 24     int cityNum;
 25     int roadNum;
 26     int myCity;
 27     int aimCity;
 28     City city[MaxSize];
 29     /* data */
 30 } Country, *ptrCountry;
 31 
 32 void getCountry(ptrCountry s);
 33 void goFind(ptrCountry C);
 34 void getAllPath(int start, vector<vector<int>> &path, vector<vector<int>> &allpath, vector<int> temperpath);
 35 
 36 int main()
 37 {
 38     //freopen("data.txt", "r", stdin);
 39     Country C;
 40     getCountry(&C);
 41     goFind(&C);
 42     return 0;
 43 }
 44 void getCountry(ptrCountry s)
 45 {
 46     int a, b, c, d;
 47     cin >> a >> b >> c >> d;
 48     s->cityNum = a;
 49     s->roadNum = b;
 50     s->myCity = c;
 51     s->aimCity = d;
 52     for (int i = 0; i < a; i++)
 53     {
 54         int t;
 55         cin >> t;
 56         s->city[i].man = t;
 57         s->city[i].firstRoad = NULL;
 58     }
 59     for (int i = 1; i <= b; i++)
 60     {
 61         int j, k, l;
 62         cin >> j >> k >> l;
 63         ptrRoad road = (ptrRoad)malloc(sizeof(Road));
 64         road->City = k;
 65         road->distance = l;
 66         road->nextRoad = s->city[j].firstRoad;
 67         s->city[j].firstRoad = road;
 68         ptrRoad road2 = (ptrRoad)malloc(sizeof(Road));
 69         road2->City = j;
 70         road2->distance = l;
 71         road2->nextRoad = s->city[k].firstRoad;
 72         s->city[k].firstRoad = road2;
 73     }
 74 }
 75 
 76 void goFind( ptrCountry C)
 77 {
 78     /*Dijstra改良算法*/
 79     int dist[C->cityNum], isVisted[C->cityNum];
 80     vector<vector<int>> path(C->cityNum);
 81     for (int i = 0; i < C->cityNum; i++)
 82     {
 83         dist[i] = INT_MAX;
 84         isVisted[i] = 0;
 85     }
 86     ptrRoad Troad = C->city[C->myCity].firstRoad;
 87     dist[C->myCity] = 0;
 88     path[C->myCity].push_back(-1);
 89     isVisted[C->myCity] = 1;
 90     while (Troad)
 91     {
 92         dist[Troad->City] = Troad->distance;
 93         path[Troad->City].push_back(C->myCity);
 94         Troad = Troad->nextRoad;
 95     }
 96     while (1)
 97     {
 98         int V = -1, min = INT_MAX - 1;
 99         for (int i = 0; i < C->cityNum; i++)
100             if ((dist[i] < min) && (!isVisted[i]))
101             {
102                 min = dist[i];
103                 V = i;
104             }
105         if (V == -1)
106             break;
107         isVisted[V] = 1;
108         ptrRoad Proad = C->city[V].firstRoad;
109         while (Proad)
110         {
111             if (!isVisted[Proad->City])
112             {
113                 if (dist[V] + Proad->distance < dist[Proad->City])
114                 {
115                     dist[Proad->City] = dist[V] + Proad->distance;
116                     path[Proad->City].clear();
117                     path[Proad->City].push_back(V);
118                 }
119                 else if (dist[V] + Proad->distance == dist[Proad->City])
120                     path[Proad->City].push_back(V);
121             }
122             Proad = Proad->nextRoad;
123         }
124     }
125     vector<vector<int>> allpath;
126     vector<int> temperpath;
127     getAllPath(C->aimCity,path, allpath,temperpath);
128     int maxMan = 0;
129     auto maxPath = allpath[0];
130     for (auto i : allpath)
131     {
132         int sum = 0;
133         for (auto j : i)
134             sum += C->city[j].man;
135         if (sum > maxMan)
136         {
137             maxMan = sum;
138             maxPath.swap(i);
139         }
140     }
141 
142     cout << allpath.size() << " " << maxMan << endl;
143 
144     for (auto i = maxPath.rbegin(); i != maxPath.rend(); i++)
145     {
146         cout << *i;
147         if((*i)!=maxPath[0])
148             cout << " ";
149     }
150     return;
151 }
152 
153 void getAllPath(int start, vector<vector<int>> &path, vector<vector<int>> &allpath, vector<int> temperpath)
154 {
155     temperpath.push_back(start);
156     if (path[start][0] == -1)
157     {
158         allpath.push_back(temperpath);
159         return;
160     }
161     for (int i = 0; i < (int)path[start].size(); i++)
162         getAllPath(path[start][i], path, allpath, temperpath);
163 }