[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。。。
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
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
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
时间： 2024-10-31 00:41:12