[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 = 200010;

priority_queue<pii, vector<pii>, greater<pii> > pq;
vector<E> e[maxn];
vector<int> path;
ll d[3][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] = 0;
    pq.push(pii(0, 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 = 0; 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<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
            num*=10,num+=in-'0';
    }
    if(IsN) num=-num;
    return true;
}

int main() {
    // freopen("in", "r", stdin);
    int T, _ = 1;
    scan_d(T);
    while(T--) {
        for(int i = 0; i < maxn; i++) e[i].clear();
        path.clear();
        scan_d(n); scan_d(m);
        for(int i = 1; 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, 0));
            }
        }
        dijkstra<0>(1); dijkstra<1>(n);
        ll cur = inf;
        for(int i = 1; i <= n; i++) {
            cur = min(cur, max(d[0][i], d[1][i]));
        }
        for(int i = 1; i <= n; i++) {
            if(cur == max(d[0][i], d[1][i])) {
                path.push_back(i);
            }
        }
        printf("Case #%d: ", _++);
        if(cur == inf) {
            printf("Evil John
");
            continue;
        }
        printf("%I64d
", cur);
        for(int i = 0; i < path.size(); i++) {
            printf("%d", path[i]);
            if(i == path.size() - 1) printf("
");
            else printf(" ");
        }
    }
    return 0;
}