[HDOJ5521]Meeting（最短路）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5521

给n个点，m个块。块内点到点之间话费的时间ti。两个人分别从点1和点n出发，问两人是否可以相遇，并求出相遇最短时间和路径，路径按照字典序输出。

这题的难点在于处理块内的点到块外点的关系。我们可以添加一个“集合点”，此点到集合内各点距离为w，点到此集合的距离为0建图。

从1和n分别做一次最短路，找到每一个距离最长的点（即为相遇点），记下长度再枚举两个结果，看一共多少个相遇点

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>

 using namespace std;

 typedef long long ll;
 typedef pair<int, int> pii;
 typedef struct E {
     int w;
     int v;
     E() {}
     E(int vv, int ww) : v(vv), w(ww) {}
 }E;

 const int inf = 0x7f7f7f7f;
 const int maxn = ;

 priority_queue<pii, vector<pii>, greater<pii> > pq;
 vector<E> e[maxn];
 vector<int> path;
 ll d[][maxn];
 int n, m, u, v, w, k;

 template<int cho>
 void dijkstra(int s) {
     memset(d[cho], inf, sizeof(d[cho]));
     while(!pq.empty()) pq.pop();
     d[cho][s] = ;
     pq.push(pii(, s));
     while(!pq.empty()) {
         pii cur = pq.top(); pq.pop();
         w = cur.first;
         v = cur.second;
         if(d[cho][v] < w) continue;
         for(int i = ; i < e[v].size(); i++) {
             if(d[cho][e[v][i].v] > d[cho][v] + e[v][i].w) {
                 d[cho][e[v][i].v] = d[cho][v] + e[v][i].w;
                 pq.push(pii(d[cho][e[v][i].v], e[v][i].v));
             }
         }
     }
 }

 inline bool scan_d(int &num) {
     char in;bool IsN=false;
     in=getchar();
     if(in==EOF) return false;
     while(in!='-'&&(in<''||in>'')) in=getchar();
     if(in=='-'){ IsN=true;num=;}
     else num=in-'';
     while(in=getchar(),in>=''&&in<=''){
             num*=,num+=in-'';
     }
     if(IsN) num=-num;
     return true;
 }

 int main() {
     // freopen("in", "r", stdin);
     int T, _ = ;
     scan_d(T);
     while(T--) {
         for(int i = ; i < maxn; i++) e[i].clear();
         path.clear();
         scan_d(n); scan_d(m);
         for(int i = ; i <= m; i++) {
             scan_d(w); scan_d(k);
             while(k--) {
                 scanf("%d", &v);
                 e[n+i].push_back(E(v, w));
                 e[v].push_back(E(n+i, ));
             }
         }
         dijkstra<>(); dijkstra<>(n);
         ll cur = inf;
         for(int i = ; i <= n; i++) {
             cur = min(cur, max(d[][i], d[][i]));
         }
         for(int i = ; i <= n; i++) {
             if(cur == max(d[][i], d[][i])) {
                 path.push_back(i);
             }
         }
         printf("Case #%d: ", _++);
         if(cur == inf) {
             printf("Evil John\n");
             continue;
         }
         printf("%I64d\n", cur);
         for(int i = ; i < path.size(); i++) {
             printf("%d", path[i]);
             if(i == path.size() - ) printf("\n");
             else printf(" ");
         }
     }
     return ;
 }