HDU 6136 贪心+优先队列+并查集优化

题意：

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6136
一个长度为L的环形赛道上有n辆赛车，每个赛车有一个初始位置和一个顺时针或逆时针的速度，第i个赛车重量为i，当两个赛车相遇之后重量小的会被毁掉，重量大将状态不变，继续前进，问经过多长时间之后，只剩下一辆车。

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思路：

贪心，可以想到一个明显的结论，所有车里最先相撞的一定是某两个相邻的赛车，那么可以将所有相邻的赛车相撞需要的时间保存在优先队列里，每次选出时间最小的一对，这两辆车会最先相撞，小的毁掉，剩下一个大的，将大的与其左右两边还存在的赛车构成两对加入队列，一直到只剩一辆车结束。
大体思路出来了，但是在每次撞击之后添加新的赛车对的过程，需要利用并查集优化，找到左右两边还存活的最近的赛车。对于规定的分数输出，只需要在最后一对时候保存分数，其余时候都用double即可。

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代码：

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int MAXN = 1e5 + 10;struct node {    int u, v;    double t;    bool operator < (const node &rhs) const {        return t - rhs.t > 1e-6;    }};struct NODE {    int d, v, id;    bool operator < (const NODE &rhs) const {        return d < rhs.d;    }} p[MAXN];int n, len;double getTime(NODE a, NODE b) {    if (a.d > b.d) swap(a, b);    if (a.v >= 0 && b.v >= 0) {        if (a.v >= b.v) return 1.0 * (b.d - a.d) / (a.v - b.v);        else return 1.0 * (len - (b.d - a.d)) / (b.v - a.v);    }    else if (a.v < 0 && b.v >= 0) {        return 1.0 * (len - (b.d - a.d)) / (b.v - a.v);    }    else if (a.v >= 0 && b.v < 0) {        return 1.0 * (b.d - a.d) / (a.v - b.v);    }    else {        if (a.v >= b.v) return 1.0 * (b.d - a.d) / (a.v - b.v);        else return 1.0 * (len - (b.d - a.d)) / (b.v - a.v);    }}int L[MAXN], R[MAXN];bool die[MAXN];int findL(int x) {    return !die[x] ? x : L[x] = findL(L[x]);}int findR(int x) {    return !die[x] ? x : R[x] = findR(R[x]);}int main() {    //freopen("in.txt", "r", stdin);    int T;    scanf("%d", &T);    while (T--) {        scanf("%d%d", &n, &len);        for (int i = 1; i <= n; i++) {            scanf("%d", &p[i].d);        }        for (int i = 1; i <= n; i++) {            scanf("%d", &p[i].v);            p[i].id = i;        }        sort (p + 1, p + 1 + n);        L[1] = n; R[n] = 1;        for (int i = 2; i <= n; i++) L[i] = i - 1;        for (int i = 1; i < n; i++) R[i] = i + 1;        priority_queue <node> que;        memset(die, false, sizeof(die));        node now, next;        for (int i = 1; i < n; i++) {            now.u = i; now.v = i + 1;            now.t = getTime(p[i], p[i + 1]);            que.push(now);        }        if (n > 2) {            now.u = 1; now.v = n;            now.t = getTime(p[1], p[n]);            que.push(now);        }        pair <int, int> ans;        while (!que.empty()) {            now = que.top(); que.pop();            int u = now.u, v = now.v;            if (die[u] || die[v]) continue;            if (p[u].id > p[v].id) swap(u, v);            ans = make_pair(u, v);            die[u] = true;            next.v = v;            next.u = findR(R[v]);            if (next.u != 0 && next.u != next.v) {                next.t = getTime(p[next.u], p[next.v]);                que.push(next);            }            next.u = findL(L[v]);            if (next.u != 0 && next.u != next.v) {                next.t = getTime(p[next.u], p[next.v]);                que.push(next);            }        }        NODE a = p[ans.first], b = p[ans.second];        LL x, y;        if (a.d > b.d) swap(a, b);        if (a.v >= 0 && b.v >= 0) {            if (a.v >= b.v) x = b.d - a.d, y = a.v - b.v;            else x = len - (b.d - a.d), y = b.v - a.v;        }        else if (a.v < 0 && b.v >= 0) {            x = len - (b.d - a.d), y = b.v - a.v;        }        else if (a.v >= 0 && b.v < 0) {            x = b.d - a.d, y = a.v - b.v;        }        else {            if (a.v >= b.v) x = b.d - a.d, y = a.v - b.v;            else x = len - (b.d - a.d), y = b.v - a.v;        }        LL g = __gcd(x, y);        x /= g; y /= g;        printf("%I64d/%I64d\n", x, y);    }    return 0;}