NOIP模拟（10.26）T2 做运动

做运动

题目背景：

10.26 NOIP模拟T2

分析：并查集 + 最短路dijkstra

 

考场上卡常数卡到想打人，出了点稍微构造过的数据······然后在本机上1.1s ~ 1.2s死都卡不进1s，然而，貌似评测姬还是比较给力的，我还是过掉了，虽然有两个点卡的不行······直接说标算吧，首先因为第一优先的条件是最高温度最低，那么我们将所有的边按照温度从小到大排序，然后依次加入，然后并查集维护连通性，直到s，t在同一个连通块，将温度小于等于这个温度的所有边加入，然后用这些加入的边跑一发最短路就好了。不过真的卡常到死啊，不开O2T飞飞啊·······

Source：

/*created by scarlyw*/#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <algorithm>#include <cctype>#include <set>#include <map>#include <vector>#include <queue>#include <ctime>inline char read() {static const int IN_LEN = 1024 * 1024;static char buf[IN_LEN], *s, *t;if (s == t) {t = (s = buf) + fread(buf, 1, IN_LEN, stdin);if (s == t) return -1;}return *s++;}///*template<class T>inline void R(T &x) {static bool iosig;static char c;for (iosig = false, c = read(); !isdigit(c); c = read()) {if (c == -1) return ;if (c == '-') iosig = true;}for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;}//*/const int OUT_LEN = 1024 * 1024;char obuf[OUT_LEN], *oh = obuf;inline void write_char(char c) {if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;*oh++ = c;}template<class T>inline void W(T x) {static int buf[30], cnt;if (x == 0) write_char('0');else {if (x < 0) write_char('-'), x = -x;for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;while (cnt) write_char(buf[cnt--]);}}inline void flush() {fwrite(obuf, 1, oh - obuf, stdout);}/*template<class T>inline void R(T &x) {static bool iosig;static char c;for (iosig = false, c = getchar(); !isdigit(c); c = getchar()) {if (c == -1) return ;if (c == '-') iosig = true;}for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;}//*/const int MAXN = 500000 + 10;const int MAXM = 1000000 + 10;const long long INF = 1000000010000000000;int n, m, x, s, t;struct edges {int from, to, t;long long w;inline bool operator < (const edges &a) const {return t < a.t;}} e[MAXM];struct node {int to;long long w;node(int to = 0, long long w = 0) : to(to), w(w) {};inline bool operator < (const node &a) const {return w > a.w;}} ;std::vector<node> edge[MAXN];inline void add_edge(int x, int y, long long w) {edge[x].push_back(node(y, w)), edge[y].push_back(node(x, w));}int father[MAXN], rk[MAXN];inline int get_father(int x) {return (father[x] == x) ? (x) : (father[x] = get_father(father[x]));}inline void unite(int x, int y) {int fa1 = get_father(x), fa2 = get_father(y);if (fa1 != fa2) {(rk[fa1] == rk[fa2]) ? (father[fa1] = fa2, rk[fa2]++) : (rk[fa1] < rk[fa2] ? father[fa1] = fa2 : father[fa2] = fa1);}}inline void read_in() {R(n), R(m);for (int i = 1; i <= n; ++i) father[i] = i, rk[i] = 0;for (int i = 1; i <= m; ++i) {R(e[i].from), R(e[i].to), R(e[i].t);R(x), e[i].w = (long long)x * (long long)e[i].t;}R(s), R(t);}inline void solve() {int ans;std::sort(e + 1, e + m + 1);for (register int i = 1; i <= m; ++i) {unite(e[i].from, e[i].to), add_edge(e[i].from, e[i].to, e[i].w);if (get_father(s) == get_father(t)) {ans = e[i].t;while (i < m && e[i + 1].t == e[i].t) ++i, add_edge(e[i].from, e[i].to, e[i].w);break ;}}static long long dis[MAXN];static bool vis[MAXN];std::priority_queue<node> q;for (register int i = 1; i <= n; ++i) dis[i] = INF;dis[s] = 0, q.push(node(s, 0));while (!q.empty()) {while (!q.empty() && vis[q.top().to]) q.pop();if (q.empty()) break ;int cur = q.top().to;if (cur == t) break ;q.pop(), vis[cur] = true;for (register int p = edge[cur].size() - 1; p >= 0; --p) {node *e = &edge[cur][p];if (!vis[e->to] && dis[e->to] > dis[cur] + e->w)dis[e->to] = dis[cur] + e->w, q.push(node(e->to, dis[e->to]));}}printf("%d %lld", ans, dis[t]);}int main() {//freopen("running.in", "r", stdin);//freopen("running.out", "w", stdout);read_in();solve();return 0;}