Codeforces Round#413 Problem A - C

Posted 2020-09-18

[写在前面感(乱)叹(七)人(八)生(糟)的话]

　　本想借此机会一口气玩到蓝名，结果，A题写炸_{(少判了一种情况)}，C题写炸_{(辜负了我5分钟狂敲出来的线段树)}，结果又掉Rating...内心好绝望。。。

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Problem#A Carrot Cakes

vjudge链接[here]

（偷个懒,cf链接就不给了）

　　题目大意是说，烤面包，给出一段时间内可以考的面包数，建第二个炉子的时间，需要达到的面包数，问建炉子是否合理。

　　玄学 & 智商题，可能是因为我智商不够，所以在我决定休息的时候被hank掉了。。。

　　输NO的情况大概是当炉子建好后，考完第一波的时间，任务已经或刚好完成。

Code

 1 /**
 2  * Codeforces
 3  * Problem#799A
 4  * Accepted
 5  * Time:15ms
 6  * Memory:8k
 7  */
 8 #include<iostream>
 9 #include<fstream> 
10 #include<sstream>
11 #include<algorithm>
12 #include<cstdio>
13 #include<cstring>
14 #include<cstdlib>
15 #include<cctype>
16 #include<cmath>
17 #include<ctime>
18 #include<map>
19 #include<stack>
20 #include<set>
21 #include<queue>
22 #include<vector>
23 using namespace std;
24 typedef bool boolean;
25 #define inf 0xfffffff
26 #define smin(a, b) (a) = min((a), (b))
27 #define smax(a, b) (a) = max((a), (b))
28 template<typename T>
29 inline boolean readInteger(T& u) {
30     char x;
31     int aFlag = 1;
32     while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1);
33     if(x == -1)    {
34         ungetc(x, stdin);
35         return false;
36     }
37     if(x == ‘-‘) {
38         aFlag = -1;
39         x = getchar();
40     }
41     for(u = x - ‘0‘; isdigit((x = getchar())); u = u * 10 + x - ‘0‘);
42     u *= aFlag;
43     ungetc(x, stdin);
44     return true;
45 }
46                                                                         
47 int n, t, k, d;
48     
49 inline void work() {
50     readInteger(n);
51     readInteger(t);
52     readInteger(k);
53     readInteger(d);
54     int turns = d / t + 1;
55     if(turns * k > n)
56         puts("NO");
57     else if(turns * k == n && (d == t || turns * t <= d + t))
58         puts("NO");
59     else
60         puts("YES");
61 }
62         
63 int main() {
64     work();
65     return 0;
66 }

Problem#A

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Problem#B T-shirt buying

vjudge链接[here]

　　题目大意是说，每件衣服有两面，每面有个颜色，有些人要来买，他只买存在一面有他喜欢的颜色且价格最低的一个，输出每个人付的钱，没有买到输出-1。

　　用个set，依题意乱搞就好了。(每场比赛貌似都是b题稳AC)

Code

  1 /**
  2  * Codeforces
  3  * Problem#799B
  4  * Accepted
  5  * Time:296ms
  6  * Memory:14964k
  7  */ 
  8 #include<iostream>
  9 #include<fstream> 
 10 #include<sstream>
 11 #include<algorithm>
 12 #include<cstdio>
 13 #include<cstring>
 14 #include<cstdlib>
 15 #include<cctype>
 16 #include<cmath>
 17 #include<ctime>
 18 #include<map>
 19 #include<stack>
 20 #include<set>
 21 #include<queue>
 22 #include<vector>
 23 using namespace std;
 24 typedef bool boolean;
 25 #define inf 0xfffffff
 26 #define smin(a, b) (a) = min((a), (b))
 27 #define smax(a, b) (a) = max((a), (b))
 28 template<typename T>
 29 inline boolean readInteger(T& u) {
 30     char x;
 31     int aFlag = 1;
 32     while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1);
 33     if(x == -1)    {
 34         ungetc(x, stdin);
 35         return false;
 36     }
 37     if(x == ‘-‘) {
 38         aFlag = -1;
 39         x = getchar();
 40     }
 41     for(u = x - ‘0‘; isdigit((x = getchar())); u = u * 10 + x - ‘0‘);
 42     u *= aFlag;
 43     ungetc(x, stdin);
 44     return true;
 45 }
 46 
 47 typedef class Tshirt {
 48     public:
 49         int p;
 50         int id;
 51         
 52         Tshirt() {        }
 53         Tshirt(int p, int id):p(p), id(id) {    }
 54         
 55         boolean operator < (Tshirt b) const {
 56             if(p != b.p) return p < b.p;
 57             return id < b.id;
 58         }
 59 }Tshirt;
 60 
 61 int n, m;
 62 int* prices;
 63 int *color1, *color2;
 64 
 65 multiset<Tshirt> ps[3];
 66 
 67 inline void init() {
 68     readInteger(n);
 69     prices = new int[(const int)(n + 1)];
 70     color1 = new int[(const int)(n + 1)];
 71     color2 = new int[(const int)(n + 1)];
 72     for(int i = 1; i <= n; i++)
 73         readInteger(prices[i]);
 74     for(int i = 1; i <= n; i++) {
 75         readInteger(color1[i]);
 76         ps[color1[i] - 1].insert(Tshirt(prices[i], i));
 77     }
 78     for(int i = 1; i <= n; i++) {
 79         readInteger(color2[i]);
 80         if(color1[i] != color2[i])
 81             ps[color2[i] - 1].insert(Tshirt(prices[i], i));
 82     }
 83 }
 84 
 85 inline void solve() {
 86     int c;
 87     readInteger(m);
 88     while(m--) {
 89         readInteger(c);
 90         c--;
 91         if(ps[c].size() == 0) {
 92             printf("-1 ");
 93         } else {
 94             multiset<Tshirt>::iterator t = ps[c].begin();
 95             printf("%d ", t->p);
 96             if(color1[t->id] != c + 1) {
 97                 ps[color1[t->id] - 1].erase(ps[color1[t->id] - 1].find(Tshirt(t->p, t->id)));
 98             } else if (color2[t->id] != c + 1) {
 99                 ps[color2[t->id] - 1].erase(ps[color2[t->id] - 1].find(*t));
100             }
101             ps[c].erase(t);
102         }
103     }
104 }
105 
106 int main() {
107     init();
108     solve();
109     return 0;
110 }

Problem#B

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Problem#C Fountains

vjudge链接[here]

　　C题思路不是很难，很好想，结果写炸了。。。

　　第一种情况，买两个不同购买方式的物品，贪心就好了。

　　第二种情况，买两个购买方式相同的物品，先按价格从小到大拍一道序，然后记录一个能延伸到的一个最远的物品r,每次i++时要用while循环去更新r，接着剩下的事交给线段树查最大值就好了。两种购买方式的物品都做一下就好了。

　　当然，写炸的缘故

　　1.自己坑自己。。sort的cmpare函数改了后没有改for循环的顺序(第一种情况比较懒，开始就写了就不想改了)

　　2.少特判当某种购买方式的物品不存在的情况，然后建树就挂了，MLE。

Code

  1 /**
  2  * Codeforces
  3  * Problem#799C
  4  * Accepted
  5  * Time:62ms
  6  * Memory:6300k
  7  */
  8 #include<iostream>
  9 #include<fstream> 
 10 #include<sstream>
 11 #include<algorithm>
 12 #include<cstdio>
 13 #include<cstring>
 14 #include<cstdlib>
 15 #include<cctype>
 16 #include<cmath>
 17 #include<ctime>
 18 #include<map>
 19 #include<stack>
 20 #include<set>
 21 #include<queue>
 22 #include<vector>
 23 using namespace std;
 24 typedef bool boolean;
 25 #define inf 0xfffffff
 26 #define smin(a, b) (a) = min((a), (b))
 27 #define smax(a, b) (a) = max((a), (b))
 28 template<typename T>
 29 inline boolean readInteger(T& u) {
 30     char x;
 31     int aFlag = 1;
 32     while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1);
 33     if(x == -1)    {
 34         ungetc(x, stdin);
 35         return false;
 36     }
 37     if(x == ‘-‘) {
 38         aFlag = -1;
 39         x = getchar();
 40     }
 41     for(u = x - ‘0‘; isdigit((x = getchar())); u = u * 10 + x - ‘0‘);
 42     u *= aFlag;
 43     ungetc(x, stdin);
 44     return true;
 45 }
 46 
 47 typedef class MyPair {
 48     public:
 49         int c;
 50         int p;
 51         
 52         MyPair(int c = 0, int p = 0):c(c), p(p) {    }
 53 }MyPair;
 54 
 55 typedef class SegTreeNode {
 56     public:
 57         int val;
 58         SegTreeNode* l, *r;
 59         
 60         SegTreeNode():val(0), l(NULL), r(NULL) {        }
 61         
 62         inline void pushUp() {
 63             val = max(l->val, r->val);
 64         }
 65 }SegTreeNode;
 66 
 67 typedef class SegTree {
 68     public:
 69         SegTreeNode* root;
 70         
 71         SegTree() {        }
 72         SegTree(int n, MyPair*& lis) {
 73             build(root, 1, n, lis);
 74         }
 75         
 76         void build(SegTreeNode*& node, int l, int r, MyPair*& lis) {
 77             node = new SegTreeNode();
 78             if(l == r) {
 79                 node->val = lis[l].c;
 80                 return;
 81             }
 82             int mid = (l + r) >> 1;
 83             build(node->l, l, mid, lis);
 84             build(node->r, mid + 1, r, lis);
 85             node->pushUp();
 86         }
 87         
 88         int query(SegTreeNode*& node, int l, int r, int ql, int qr) {
 89             if(ql > qr)    return 0;
 90             if(l == ql && r == qr) {
 91                 return node->val;
 92             }
 93             int mid = (l + r) >> 1;
 94             if(qr <= mid)    return query(node->l, l, mid, ql, qr);
 95             if(ql > mid)    return query(node->r, mid + 1, r, ql, qr);
 96             int a = query(node->l, l, mid, ql, mid);
 97             int b = query(node->r, mid + 1, r, mid + 1, qr);
 98             return max(a, b);
 99         }
100         
101         void clear(SegTreeNode*& node) {
102             if(node->l)    clear(node->l);
103             if(node->r)    clear(node->r);
104             delete node;
105         }
106 }SegTree;
107 
108 int n;
109 int C, D;
110 MyPair *cs, *ds;
111 int cnc = 0, cnd = 0;
112 int bc = -1, bd = -1;
113 SegTree stc, stdd;
114 
115 inline void init() {
116     readInteger(n);
117     readInteger(C);
118     readInteger(D);
119     cs = new MyPair[(const int)(n + 1)];
120     ds = new MyPair[(const int)(n + 1)];
121     int a, b;
122     char x;
123     for(int i = 1; i <= n; i++) {
124         readInteger(a);
125         readInteger(b);
126         getchar(); x = getchar();
127         if(x == ‘C‘ && b <= C)
128             cs[++cnc] = MyPair(a, b);
129         else if(x == ‘D‘ && b <= D)
130             ds[++cnd] = MyPair(a, b);
131     }
132 }
133 
134 boolean cmp1(const MyPair& a, const MyPair& b) {
135     if(a.c != b.c) return a.c > b.c;
136     return a.p < b.p;
137 }
138 
139 boolean cmp2(const MyPair& a, const MyPair& b) {
140     if(a.p != b.p) return a.p < b.p;
141     return a.c > b.c;
142 }
143 
144 int res = 0;
145 
146 inline void solve() {
147     if(cnc > 0)
148         sort(cs + 1, cs + cnc + 1, cmp1);
149     if(cnd > 0)
150         sort(ds + 1, ds + cnd + 1, cmp1);
151     for(int i = 1; i <= cnc; i++)
152         if(cs[i].p <= C) {
153             bc = cs[i].c;
154             break;
155         }
156     for(int i = 1; i <= cnd; i++)
157         if(ds[i].p <= D) {
158             bd = ds[i].c;
159             break;
160         }
161     if(bc != -1 && bd != -1)
162         res = bc + bd;
163         
164     if(cnc > 0)
165         sort(cs + 1, cs + cnc + 1, cmp2);
166     if(cnd > 0)
167         sort(ds + 1, ds + cnd + 1, cmp2);
168     int r = cnc;
169     if(cnc) {
170         stc = SegTree(cnc, cs);
171         for(int i = 1; i <= r; i++) {
172             int a0 = 0, a1 = 0;
173             while(cs[i].p + cs[r].p > C)    r--;
174             if(i > r)   break;
175             if(i > 1)
176                 a0 = stc.query(stc.root, 1, cnc, 1, i - 1);
177             if(i < cnc)
178                 a1 = stc.query(stc.root, 1, cnc, i + 1, r);
179             if(a0 == 0 && a1 == 0)    continue;
180             smax(res, max(a0, a1) + cs[i].c);
181         }
182     }
183     
184     r = cnd;
185     if(cnd) {
186     stdd = SegTree(cnd, ds);
187         for(int i = 1; i <= r; i++) {
188             int a0 = 0, a1 = 0;
189             while(ds[i].p + ds[r].p > D)    r--;
190             if(i > r)    break;
191             if(i > 1)
192                 a0 = stdd.query(stdd.root, 1, cnd, 1, i - 1);
193             if(i < cnd)
194                 a1 = stdd.query(stdd.root, 1, cnd, i + 1, r);
195             if(a0 == 0 && a1 == 0)    continue;
196             smax(res, max(a0, a1) + ds[i].c);
197         }
198     }
199     printf("%d", res);
200 }
201 
202 int main() {
203     init();
204     solve();
205     return 0;
206 }

Problem#C