概述
包含一号线、二号线全部,还有两小段
数据来自北京地铁官网
资源链接
https://download.csdn.net/download/renzemingcsdn/44352688
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <map>
#include <string>
using namespace std;
const int MAX_N = 99;
const int INF = 999999;
// 各数组都从下标1开始
int dst[MAX_N]; // 表示当前点到源点的最短路径长度
int prv[MAX_N]; // 记录当前点的前一个结点
int c[MAX_N][MAX_N]; // 记录图的两点间路径长度
int n=46, line=0; // 图的结点数和路径数
void Dijkstra(int n, int v)
{
bool vis[MAX_N]; // 判断是否已存入该点到S集合中
for (int i = 1; i <= n; ++i)
{
dst[i] = c[v][i];
vis[i] = 0; // 初始都未用过该点
if (dst[i] == INF)
prv[i] = 0;
else
prv[i] = v;
}
dst[v] = 0;
vis[v] = 1;
// 依次将未放入S集合的结点中,取dst[]最小值的结点,放入结合S中
// 一旦S包含了所有V中顶点,dst就记录了从源点到所有其他顶点之间的最短路径长度
// 注意是从第二个节点开始,第一个为源点
for (int i = 2; i <= n; ++i)
{
int tmp = INF;
int u = v;
// 找出当前未使用的点j的dst[j]最小值
for (int j = 1; j <= n; ++j)
if ((!vis[j]) && dst[j] < tmp)
{
u = j; // u保存当前邻接点中距离最小的点的号码
tmp = dst[j];
}
vis[u] = 1; // 表示u点已存入S集合中
// 更新dst
for (int j = 1; j <= n; ++j)
if ((!vis[j]) && c[u][j] < INF)
{
int newdst = dst[u] + c[u][j];
if (newdst < dst[j])
{
dst[j] = newdst;
prv[j] = u;
}
}
}
}
void c_init() {
for (int i = 0; i < MAX_N; i++) {
dst[i] = INF;
for (int j = 0; j < MAX_N; j++) {
c[i][j] = INF;
}
}
c[1][2] = c[2][1] = 2606;
c[2][3] = c[3][2] = 1921;
c[3][4] = c[4][3] = 1953;
c[4][5] = c[5][4] = 1479;
c[5][6] = c[6][5] = 1810;
c[6][7] = c[7][6] = 1778;
c[7][8] = c[8][7] = 1313;
c[8][9] = c[9][8] = 1172;
c[9][10] = c[10][9] = 1166;
c[10][11] = c[11][10] = 1291;
c[11][12] = c[12][11] = 424;
c[12][13] = c[13][12] = 1590;
c[13][14] = c[14][13] = 1217;
c[14][15] = c[15][14] = 925;
c[15][16] = c[16][15] = 852;
c[16][17] = c[17][16] = 774;
c[17][18] = c[18][17] = 1230;
c[18][19] = c[19][18] = 1372;
c[19][20] = c[20][19] = 790;
c[20][21] = c[21][20] = 1385;
c[21][22] = c[22][21] = 1673;
c[22][23] = c[23][22] = 1714;
c[12][24] = c[24][12] = 1832;
c[24][25] = c[25][24] = 960;
c[25][26] = c[25][26] = 909;
c[26][27] = c[27][26] = 1899;
c[27][28] = c[28][27] = 1766;
c[28][29] = c[29][28] = 1237;
c[29][30] = c[30][29] = 794;
c[30][31] = c[31][30] = 2228;
c[31][32] = c[32][31] = 824;
c[32][33] = c[33][32] = 1024;
c[33][18] = c[18][33] = 1763;
c[34][18] = c[18][34] = 945;
c[34][35] = c[35][34] = 1023;
c[35][36] = c[36][35] = 1034;
c[36][37] = c[37][36] = 1171;
c[37][38] = c[38][37] = 851;
c[38][39] = c[39][38] = 939;
c[12][39] = c[39][12] = 1234;
c[26][40] = c[40][26] = 1025;
c[40][41] = c[41][40] = 1100;
c[41][42] = c[42][41] = 1100;
c[42][43] = c[43][42] = 869;
c[43][13] = c[13][43] = 1011;
c[13][38] = c[38][13] = 815;
c[30][44] = c[44][30] = 866;
c[44][45] = c[45][44] = 791;
c[45][46] = c[46][45] = 1016;
c[46][47] = c[47][46] = 848;
c[47][17] = c[17][47] = 945;
c[17][35] = c[35][17] = 821;
}
map<string, int> mp_stoi = {
{"pingguoyuan",1},
{"gucheng",2},
{"bajiao amusement park",3},
{"babaoshan",4},
{"yuquanlu",5},
{"wukesong",6},
{"wanshoulu",7},
{"gongzhufen",8},
{"militayr museum",9},
{"muxidi",10},
{"nanlishilu",11},
{"fuxingmen",12},
{"xidan",13},
{"tian anmen west",14},
{"tian anmen east",15},
{"wangfujing",16},
{"dongdan",17},
{"jianguomen",18},
{"yonganli",19},
{"guomao",20},
{"dawanglu",21},
{"sihui",22},
{"sihui east",23},
{"fuchengmen",24},
{"chegongzhuang",25},
{"xizhimen",26},
{"jishuitan",27},
{"guloudajie",28},
{"andingmen",29},
{"yonghegong",30},
{"dongzhimen",31},
{"dongshisitiao",32},
{"chaoyangmen",33},
{"beijingzhan",34},
{"chongwenmen",35},
{"qianmen",36},
{"hepingmen",37},
{"xuanwumen",38},
{"changchunjie",39},
{"xinjiekou",40},
{"pinganli",41},
{"xisi",42},
{"lingjinghutong",43},
{"beixinqiao",44},
{"zhangzizhognlu",45},
{"dongsi",46},
{"dengshikou",47},
};
int main()
{
c_init();
string s;
getline(cin, s);
int i;
for (i = 0; i < s.size(); i++) {
if (s[i] == ';') {
break;
}
}
string s1, s2;
s1=s.substr(0,i);
s2 = s.substr(i + 1, s.size()-1);
int start, end;
start = mp_stoi[s1];
end = mp_stoi[s2];
if (start == 0 || end == 0) {
return -1;
}
Dijkstra(47, start);
int l = dst[end];
if (l <= 6000) {
cout << 3 << endl;
}
else if (l <= 12000) {
cout << 4 << endl;
}
else if (l <= 22000) {
cout << 5 << endl;
}
else if (l <= 32000) {
cout << 6 << endl;
}
else {
cout << (6 + (l - 32) % 20 + 1) << endl;
}
}
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