HDU 4302 Holedox Eating（线段树/优先队列/multiset）

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题意：一条线段，在某些时刻上面的某些点会随机出现蛋糕。某种动物最初在0位置，在该动物想吃蛋时，它会跑到离他最近的蛋糕那里；如果有两个蛋糕与它的距离相同，就按照上次前进的方向走。输出它走过的总路程。
其实就是个水题，然而坑了一天，交了两页的记录 - - 写个博客纪念下
赛上用线段树区间最值过的，赛后看题解提到也可以线段树 + 二分做，又听说了 pq 和 multiset 的版本，就写了一发 pq，结果有个地方觉得写得不对但还是莫名AC了，觉得奇怪就坑了一下午 - - 还好最后终于找到了原因，并且用下面的数据叉掉了集训队里一个汉子的AC代码……这里output应该是26~
关于这个case和这个隐蔽修改方向导致WA的问题的解释以及multiset版本的代码，详见 lwt学长哒博客
总之，写代码前一定要思考清楚，要写出美观不易错的代码~

1  20 13  0 4  1  0 3  1  0 3  1  0 1  0 5  0 20  1  1  1  1

线段树区间最值
网上思路较多，就不细说了 - -

//赛上的线段树区间最值版#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <map>#include <queue>#include <string>#define ALL(v) v.begin(), v.end()#define CLR(array) memset(array, 0, sizeof(array))#define DEBUG cout << "bug "#define Fr first#define FOR(i, n) for(int i = 0; i < n; i++)#define MEM(array) memset(array, 0x3f, sizeof(array))#define ls id << 1#define lson id << 1, l, m#define OFF(array) memset(array, -1, sizeof(array))#define PR(x) cout << #x << " = " << x << "  "#define PRLN(x) cout << #x << " = " << x << endl#define rep(i, a, b) for(int i = a; i <= b; i++)#define root 1, 1, n#define rs id << 1 | 1#define rson id << 1 | 1, m + 1, r#define Rep(i, a, b) for(int i = a; i >= b; i--)#define repk(i, a, b) for(i = a; i <= b; i++)#define Repk(i, a, b) for(i = a; i >= b; i--)#define Sc second#define SZ size()using namespace std;typedef long long ll;typedef pair<int, int> P;const int inf = 0x3f3f3f3f, maxn = 1e5 + 5;struct Node {    int lc, rc, cnt; //这里因为某点上可能有多个cake，所以还记录了区间内的cake数，其实单开数组记录每个点的cake数也可以} tree[maxn << 2];int lef, rig;void build(int id, int l, int r) {    tree[id].lc = inf;    tree[id].cnt = 0;    tree[id].rc = -inf;    if(l == r)         return;    int m = l + r >> 1;    build(lson);    build(rson);}void push_up(int id) {    Node& now = tree[id];    now.lc = min(tree[ls].lc, tree[rs].lc);    now.rc = max(tree[ls].rc, tree[rs].rc);}void update(int id, int l, int r, int pos, int d) {    Node& now = tree[id];    now.cnt += d;    if(l == r) {        if(now.cnt)             now.lc = now.rc = l;        else {            now.lc = inf;            now.rc = -inf;        }        return;    }    int m = l + r >> 1;    if(pos <= m) update(lson, pos, d);    else update(rson, pos, d);    push_up(id);}void query(int id, int l, int r, int a, int b, int op) {    if(!tree[id].cnt) return;    if(a <= l && r <= b) {        if(op) rig = min(rig, tree[id].lc);        else lef = max(lef, tree[id].rc);        return;    }    int m = l + r >> 1;    if(a <= m) query(lson, a,  b, op);    if(b > m) query(rson, a, b, op);}ll solve(void) {    int n, q, a, b;    ll ans = 0;    int pre = 1, now = 1;    scanf("%d%d", &n, &q);    n++;    build(root);    while(q--) {        scanf("%d", &a);        if(a) {            int goal;            lef = -inf, rig = inf;            query(root, 1, now, 0);            query(root, now + 1, n, 1);            if(now - lef != rig - now)                goal = now - lef < rig - now ? lef : rig;            else                 goal = pre > 0 ? rig : lef;            if(abs(goal) == inf) continue;            update(root, goal, -1);            ans += abs(goal - now);            if(goal > now) pre = 1;            else if(goal < now) pre = -1;            now = goal;        } else {            scanf("%d", &b);            update(root, b + 1, 1);        }    }    return ans;}int main(void) {    int t, kase = 0;    scanf("%d", &t);    while(t--)         printf("Case %d: %I64d\n", ++kase, solve());    return 0;}

然后是二分版本哒~可以不记录区间最值，只要记录区间内cake数就可以了，但是要写两个query，不太好看。
思路是每次查询到某个区间时，如果当前区间内没有cake，直接返回。如果要查找区间最右边的值，就优先查找右子树，然后才是左子树，查找区间左值类比即可。
不过，网上虽然有人提了这个思路，但是几乎没有清晰代码，所以也贴下~

// 二分版本~// 省略头文件50行const int inf = 0x3f3f3f3f, maxn = 1e5 + 5;int sum[maxn << 2];void build(int id, int l, int r) {    sum[id] = 0;    if(l == r) return;    int m = l + r >> 1;    build(lson);    build(rson);}void update(int id, int l, int r, int pos, int d) {    sum[id] += d;    if(l == r) return;    int m = l + r >> 1;    if(pos <= m) update(lson, pos, d);    else update(rson, pos, d);}int queryl(int id, int l, int r, int a, int b) { // 查找区间最左边的cake的位置，没有则返回inf    if(l == r) return l;    int m = l + r >> 1;    int res = inf;    if(a <= m && sum[ls])         res = queryl(lson, a, b);    if(res == inf && b > m && sum[rs])        res = queryl(rson, a, b);    return res;}// 注意两个query的区别int queryr(int id, int l, int r, int a, int b) { // 查找区间最右边的cake的位置，没有则返回 -inf    if(l == r) return l;    int m = l + r >> 1;    int res = -inf;    if(b > m && sum[rs])        res =  queryr(rson, a, b);    if(res == -inf && a <= m && sum[ls])        res = queryr(lson, a, b);    return res;}ll solve(void) {    int n, q, a, b;    ll ans = 0;    int pre = 1, now = 1;    scanf("%d%d", &n, &q);    n++;    build(root);    while(q--) {        scanf("%d", &a);        if(!a) {            scanf("%d", &b);            update(root, b + 1, 1);        } else {            int goal, lef, rig;            lef = queryr(root, 1, now);            rig = queryl(root, now + 1, n);            if(now - lef != rig - now)                goal = now - lef < rig - now ? lef : rig;            else                 goal = pre > 0 ? rig : lef;            if(abs(goal) == inf) continue;            GOTO: update(root, goal, -1);            ans += abs(goal - now);            if(goal > now) pre = 1;            else if(goal < now) pre = -1;            now = goal;                }    }    return ans;}int main(void) {    int t, kase = 0;    scanf("%d", &t);    while(t--)         printf("Case %d: %I64d\n", ++kase, solve());    return 0;}

pq版本也来一发，网上代码也很多，不细说了 - -
不过说实话，用两个pq维护虽然听上去很机智，但是代码比较丑也可能是我写的不好看吧 - - 感觉 multiset 或者 map 写会更简单

// pq 版~const int inf = 0x3f3f3f3f, maxn = 1e5 + 5;ll solve(void) {    priority_queue <int> pql;    priority_queue <int, vector<int>, greater<int> > pqr;    int n, q, a, b;    ll ans = 0;    int pre = 1, now = 0;    scanf("%d%d", &n, &q);    while(q--) {        scanf("%d", &a);        if(!a) {            scanf("%d", &b);            if(b >= now) pqr.push(b);            else pql.push(b);        } else {            int choice = 1, lef = -inf, rig = inf;            if(pql.size())                lef = pql.top();            if(pqr.size())                rig = pqr.top();            if(now - lef != rig - now) {                if(now - lef < rig - now)                     choice = -1;            } else choice = pre;            if(choice > 0 && rig != inf) {                pqr.pop();                ans += rig - now;                if(rig != now)                    pre = choice;                                now = rig;            } else if(choice < 0 && lef != -inf) {                pql.pop();                ans += now - lef;                pre = choice;                now = lef;            }        }    }    return ans;}int main(void) {    int t, kase = 0;    scanf("%d", &t);    while(t--)         printf("Case %d: %I64d\n", ++kase, solve());    return 0;}

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