L3-014 周游世界

高难度dfs，解析看subway map那篇博客

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
#include<vector>
using namespace std;
const int maxn=10014;
map<int,int> line;
vector<int> g[maxn];
vector<int> path,tmp;
int transferCnt (vector<int> vi) {
    int cnt=0;
    int pre=vi[0];
    for (int i=1;i<vi.size()-1;i++) {
        if (line[pre*10000+vi[i]]!=line[vi[i]*10000+vi[i+1]])
        cnt++;
        pre=vi[i];
    }
    return cnt;
}
int minTransferCnt=1e9;
int minCnt=1e9;
int st,ed;
bool visit[maxn];
void dfs (int st) {
    tmp.push_back(st);
    visit[st]=true;
    if (st==ed) {
        if (tmp.size()<minCnt) {
            path=tmp;
            minCnt=tmp.size();
            minTransferCnt=transferCnt(tmp);
        }
        else if (tmp.size()==minCnt&&transferCnt(tmp)<minTransferCnt) {
            path=tmp;
            minTransferCnt=transferCnt(tmp);
        }
        tmp.pop_back();
        return;
    }
    for (int i=0;i<g[st].size();i++) {
    if (visit[g[st][i]]==false) {
        dfs (g[st][i]);
        visit[g[st][i]]=false;
        //tmp.pop_back();
    }
}
    tmp.pop_back();
}
int main () {
    int n,k,pre,x;
    scanf ("%d",&n);
    for (int i=1;i<=n;i++) {
        scanf ("%d %d",&k,&pre);
        for (int j=1;j<k;j++) {
            scanf ("%d",&x);
            g[x].push_back(pre);
            g[pre].push_back(x);
            line[pre*10000+x]=i;
            line[x*10000+pre]=i;
            pre=x;
        }
    }
    int q;
    scanf ("%d",&q);
    for (int i=0;i<q;i++) {
        scanf ("%d %d",&st,&ed);
        fill (visit,visit+maxn,false);
        minCnt=1e9;
        minTransferCnt=1e9;
        tmp.clear ();
        dfs (st);
        visit[st]=false;
        if (minCnt==1e9) {
            printf ("Sorry, no line is available.\n");
            continue;
        }
        printf ("%d\n",minCnt-1);
        int preL=0;
        for (int i=1;i<path.size();i++) {
            if (line[path[i-1]*10000+path[i]]!=line[path[i]*10000+path[i+1]]) {
                printf ("Go by the line of company #%d from %04d to %04d.\n",line[path[i-1]*10000+path[i]],path[preL],path[i]);
                preL=i;
            }
        }
    }
    return 0;
}