題目鏈接
2017.9.5 1:40 而且還開學了
用很大很大的數據量對拍了好久好久終於找出了錯誤
智障到哭泣
bool dir = t->ch[1]->key < t->ch[0]->key;
寫成了 bool dir = t->ch[1]->val < t->ch[0]->val;
一直RE完全不知道哪裏出了問題，只好一點一點跟別人的代碼對，很多無關緊要的地方都改成了別人寫的樣子(...)就很心累。做數據結構的題有沒有必要呢......反正比賽時也不會寫，會寫也來不及寫，寫完也來不及調，更何況還每次都有智障錯誤，。
深夜低氣壓，先扔一發代碼明天再完善博文了...。
題意
若干組
排名的依據是1. 過題數；2.
前兩個指標都一樣的話排名就相同，對於相同排名的在輸出的時候輸出其中第三個指標最優的。
法一：Treap
思路
用
爲什麼不把最後過的一題的編號也加進去呢？注意到，題目要詢問恰好第
那麼最後怎麼輸出第三個指標最優的呢？參考了 hdu 5096 ACM Rank(treap) ——yamiedie_，在節點裏面再開一個
問第
Code
#include <bits/stdc++.h>
#define maxn 20010
using namespace std;
int cnt[maxn], trial[maxn][12], last[maxn], pen[maxn], lstacc[maxn];
bool solve[maxn][12];
char s[100];
struct Sta {
int cnt, pen;
Sta(int _cnt=0, int _pen=0) : cnt(_cnt), pen(_pen) {}
bool operator < (const Sta& sta) const {
return cnt > sta.cnt || (cnt == sta.cnt && pen < sta.pen);
}
bool operator == (const Sta& sta) const {
return cnt == sta.cnt && pen == sta.pen;
}
};
struct cmp {
bool operator()(const int &a, const int &b) const{
return lstacc[a] < lstacc[b] || (lstacc[a] == lstacc[b] && a < b);
}
};
struct node {
Sta val; int key, sz, cnt;
node* ch[2];
set<int, cmp> s;
node() { sz = 0, key = INT_MAX, s.clear(); }
node(Sta sta, int id);
update() { sz = ch[0]->sz + ch[1]->sz + cnt; }
}* null = new node;
node::node(Sta sta, int id) {
val = sta, key = rand(), sz = cnt = 1;
ch[0] = ch[1] = null;
s.clear(); s.insert(id);
}
struct Treap {
node* root;
Treap() { root = null; }
void rotate(node*& t, bool d) {
node* p = t->ch[d];
t->ch[d] = p->ch[!d];
p->ch[!d] = t;
t->update(), p->update();
t = p;
}
void insert(node*& t, Sta sta, int id) {
if (t == null) { t = new node(sta, id); return; }
if (t->val == sta) {
++t->cnt, t->s.insert(id);
t->update();
return;
}
bool dir = !(sta < t->val);
insert(t->ch[dir], sta, id);
if (t->ch[dir]->key < t->key) rotate(t, dir);
else t->update();
}
void erase(node*& t, Sta sta, int id) {
if (t == null) return;
if (t->val == sta) {
if (t->cnt > 1) {
--t->cnt; t->s.erase(id);
t->update();
return;
}
bool dir = t->ch[1]->key < t->ch[0]->key;
if (t->ch[dir] == null) { delete t; t = null; return; }
if (t->ch[!dir] == null) { node* p = t->ch[dir]; delete t; t = p; return; }
rotate(t, dir);
erase(t->ch[!dir], sta, id);
t->update();
return;
}
bool dir = !(sta < t->val);
erase(t->ch[dir], sta, id);
t->update();
}
void insert(Sta sta, int id) {
insert(root, sta, id);
}
void erase(Sta sta, int id) {
erase(root, sta, id);
}
int calckth(int k) {
if (k <= 0 || k > root->sz) return -1;
bool dir;
for (node* t = root; t != null; t = t->ch[dir]) {
int num = t->ch[0]->sz;
if (k == num+1) return *(t->s.begin());
else if (k <= num) dir = 0;
else {
k -= (num + t->cnt);
if (k <= 0) return -1;
else dir = 1;
}
}
}
int find(node*& t, Sta sta) {
if (t->val == sta) return t->ch[0]->sz + 1;
bool d = !(sta < t->val);
if (d == 0) return find(t->ch[d], sta);
else return t->ch[0]->sz + t->cnt + find(t->ch[d], sta);
}
void clear(node*& t) {
if (t->ch[0] != null) clear(t->ch[0]);
if (t->ch[1] != null) clear(t->ch[1]);
delete t; t = null;
}
void clear() { clear(root); }
int find(Sta sta) { return find(root, sta); }
int size() { return root->sz; }
}* treap = new Treap;
int readint() {
char c;
while((c = getchar()) && !(c >= '0' && c <= '9'));
int ret = c - '0';
while((c = getchar()) && c >= '0' && c <= '9')
ret = ret * 10 + c - '0';
return ret;
}
char readc() {
char c;
while((c = getchar()) && !isalpha(c));
return c;
}
int n, m, kas;
void work() {
memset(cnt, 0, sizeof(cnt));
memset(pen, 0, sizeof(pen));
memset(solve, 0, sizeof(solve));
memset(last, -1, sizeof(last));
memset(trial, 0, sizeof(trial));
memset(lstacc, 0, sizeof(lstacc));
if (treap->root != null) treap->clear();
for (int i = 0; i < n; ++i) treap->insert(Sta(0, 0), i);
char op[20];
int order = 0;
while (scanf("%s", op) && op[0] != 'C') {
++order;
int k, x;
if (op[0] == 'S') {
int time = readint(), team = readint(), prob = readc()-'A', stat = readint();
if (!solve[team][prob] && (last[team] == -1 || time - last[team] >= 5)) {
if (stat != 1) { last[team] = time; ++trial[team][prob]; continue; }
Sta sta1(cnt[team], pen[team]);
treap->erase(sta1, team);
solve[team][prob] = true, ++cnt[team];
pen[team] += time + trial[team][prob] * 20;
last[team] = time;
lstacc[team] = order;
Sta sta2(cnt[team], pen[team]);
treap->insert(sta2, team);
printf("[%d][%c]\n", team, prob+'A');
}
}
else if (op[0] == 'T') {
scanf("%d", &k);
printf("%d\n", treap->calckth(k));
}
else if (op[0] == 'R') {
scanf("%d", &x);
Sta sta(cnt[x], pen[x]);
printf("%d\n", treap->find(sta));
}
}
scanf("%s", op);
printf("\n");
}
int main() {
while (scanf("%d%d", &n, &m) != EOF) work();
return 0;
}問題
- 多組數據最好要回收內存，寫在了
clear() 裏面 - 這道題的讀入一開始也出了點問題 0 0
- 輸出時每組數據之間要空行
- 應將
last[] 初始化爲−1 ，否則判斷是否相隔五分鐘的時候會出問題(…) - 第三個排序的指標不是最後一次提交的時間，而是反應在
submissionlist 上的順序… erase 時
bool dir = t->ch[1]->key < t->ch[0]->key;
寫成了bool dir = t->ch[1]->val < t->ch[0]->val;這貌似還不是第一次…。
法二：樹狀數組
等等補