[hdu 5521 Meeting] 拆边 + Dijkstra

题目链接：[hdu 5521 Meeting] 拆点 + Dijkstra
题意描述：给定N个顶点，编号从1~N，M个顶点集合， 第i个集合Si中所有顶点两两距离是ti。Bessie从顶点1出发,Elsie从顶点N出发，问他们两个最快相聚的时间，以及所以相聚的顶点的编号。注意的是，他们两人是可以同时移动的。 (2≤n≤105,1≤ti≤109,∑mi=1|Si|≤106)
解题思路：这个题目还是具有一定的技巧性的。考虑这个最短路问题，可能关键的问题是如何存边。我这里用到了拆边的技巧。也就是，对于每个集合Si我新增一个顶点，集合中所有的顶点都连接到这个新增的顶点上，边长就是ti2，考虑到浮点误差，不妨先将边长扩大两倍。最后统计的时候再除2。
分别对顶点1和顶点N求Dijkstra，那么答案就是

min{max{dij1.cost[u],dij2.cost[u]}}(1≤u≤N)

#include <bits/stdc++.h>using namespace std;#define FIN freopen("input.txt", "r", stdin)typedef __int64 LL;const int MAXN = 1e6 + 5;const int MAXE = 1e6 + 5;const LL INF = 0x3f3f3f3f3f3f3f3fLL;int T, cas;int N, M;LL ans;set<int> res;set<int>::iterator iter;struct Edge {    int v, next;    LL w;    Edge() {}    Edge(int v, LL w, int next) : v(v), w(w), next(next) {}} edges[MAXE << 1];int head[MAXN], ESZ;LL dp[MAXN];void init() {    ESZ = 0;    memset(head, -1, sizeof(head));    res.clear();}void add_edge(int u, int v, LL w) {    edges[ESZ] = Edge(v, w, head[u]);    head[u] = ESZ ++;}struct Dijkstra {    struct QNode {        int u;        LL cost;        QNode() {}        QNode(int u, LL cost) : u(u), cost(cost) {}        bool operator > (const QNode& e)const {            return cost > e.cost;        }    } cur;    bool used[MAXN];    LL cost[MAXN];    priority_queue<QNode, vector<QNode>, greater<QNode> > Q;    void init() {        memset(used, false, sizeof(used));        memset(cost, 0x3f, sizeof(cost));    }    void run(int src) {        int u, v;        LL w;        Q.push(QNode(src, cost[src] = 0));        while(!Q.empty()) {            cur = Q.top();            Q.pop();            u = cur.u;            if(used[u]) continue;            used[u] = true;            for(int i = head[u]; ~i; i = edges[i].next) {                v = edges[i].v, w = edges[i].w;                if(!used[v] && cost[v] > cost[u] + w) {                    Q.push(QNode(v, cost[v] = cost[u] + w));                }            }        }    }} dij1, dij2;int main() {#ifndef ONLINE_JUDGE    FIN;#endif // ONLINE_JUDGE    int n, u, v;    LL w;    cas = 0;    scanf("%d", &T);    while(T --) {        init();        scanf("%d %d", &N, &M);        u = N;        for(int i = 0; i < M; i++) {            scanf("%I64d %d", &w, &n);            u ++;            while(n --) {                scanf("%d", &v);                add_edge(u, v, w);                add_edge(v, u, w);            }        }        dij1.init();        dij1.run(1);        if(dij1.cost[N] == INF) {            printf("Case #%d: Evil John\n", ++cas);            continue;        }        for(u = 1; u <= N; u++) {            dp[u] = dij1.cost[u] >> 1;        }        dij2.init();        dij2.run(N);        for(u = 1; u <= N; u++) {            dp[u] = max(dp[u], dij2.cost[u] >> 1);        }        ans = INF;        for(u = 1; u <= N; u++) {            ans = min(ans, dp[u]);        }        printf("Case #%d: %I64d\n", ++cas, ans);        for(u = 1; u <= N; u++) {            if(ans != dp[u]) continue;            if(dp[u] == (dij1.cost[u] >> 1) || dp[u] == (dij2.cost[u] >> 1)) {                res.insert(u);            }        }        for(iter = res.begin(); iter != res.end(); iter++) {            if(iter != res.begin()) printf(" ");            printf("%d", *iter);        }        printf("\n");    }    return 0;}