題目:click me~
題意:給出幾條地鐵線路,查詢從起點站到終點站的經停次數最少的路線,如果經停次數相同,輸出換乘次數最少的路線。
解題思路:需要一遍DFS,DFS中要維護兩個變量:mincnt中途經停最少的次數,mintransfer需要換乘的最小次數
步驟一:計算換乘次數的方法:在line[10000][10000]的數組中保存每兩個相鄰站中間的路是幾號線。從頭到尾遍歷最終的路徑,preline是前一小段的線路編號,若當前小段的線路編號與preline不同,說明有一個換乘,累加器加1;
步驟二:計算經停次數的方法:在DFS時設置一個cnt,表示當前站是路線中的第幾個站,每次向下遍歷一層就將cnt+1;
步驟三:輸出結果的方法:和計算換乘次數的思路一樣,遍歷最終的路徑,若當前line值與preline不同,輸出一句話,用pretransfer保存上一個經停站。
code
#include<iostream>
#include<vector>
#include<unordered_map>
using namespace std;
vector<vector<int>> v(10000);
int vis[10000], mincnt, mintransfer, st, ed;
unordered_map<int, int> line;
vector<int> path, temppath;
int transfercnt(vector<int> a) {
int cnt = -1, preline = 0;
for (int i = 1;i < a.size();i++) {
if (line[a[i - 1] * 10000 + a[i]] != preline)cnt++;
preline = line[a[i - 1] * 10000 + a[i]];
}
return cnt;
}
void dfs(int node, int cnt) {
if (node == ed && (cnt < mincnt || (cnt == mincnt && transfercnt(temppath) < mintransfer))) {
mincnt = cnt;
mintransfer = transfercnt(temppath);
path = temppath;
}
if (node == ed)return;
for (int i = 0;i < v[node].size();i++) {
if (vis[v[node][i]] == 0) {
vis[v[node][i]] = 1;
temppath.push_back(v[node][i]);
dfs(v[node][i], cnt + 1);
vis[v[node][i]] = 0;
temppath.pop_back();
}
}
}
int main() {
int n, m, k, pre, temp;
cin >> n;
for (int i = 0;i < n;i++) {
cin >> m >> pre;
for (int j = 1;j < m;j++) {
cin >> temp;
v[pre].push_back(temp);
v[temp].push_back(pre);
line[pre * 10000 + temp] = line[temp * 10000 + pre] = i+1;
pre = temp;
}
}
cin >> k;
for (int i = 0;i < k;i++) {
cin >> st >> ed;
mincnt = 99999, mintransfer = 99999;
temppath.clear();
temppath.push_back(st);
vis[st] = 1;
dfs(st, 0);
vis[st] = 0;
cout << mincnt << endl;
int preline = 0, pretransfer = st;
for (int j = 1;j < path.size();j++) {
if (line[path[j - 1] * 10000 + path[j]] != preline) {
if (preline != 0)printf("Take Line#%d from %04d to %04d.\n", preline, pretransfer, path[j - 1]);
preline = line[path[j - 1] * 10000 + path[j]];
pretransfer = path[j - 1];
}
}
printf("Take Line#%d from %04d to %04d.\n", preline, pretransfer, ed);
}
system("pause");
return 0;
}