|
ID
|
Origin
|
Title
| ||
|---|---|---|---|---|
 |
31 / 51 | Problem A | HDU 4175 | Class Schedule |
|
27 / 140 | Problem B | HDU 4193 | Non-negative Partial Sums |
| 5 / 17 | Problem C | HDU 4212 | Shut the Box | |
| 3 / 9 | Problem D | HDU 4216 | Computational Geometry? | |
| 2 / 3 | Problem E | HDU 4219 | Randomization? | |
| Problem F | HDU 4234 | Moving Points | ||
| 9 / 17 | Problem G | HDU 4243 | Maximum in the Cycle of 1 | |
| Problem H | HDU 4253 | Two Famous Companies | ||
|
5 / 25 | Problem I | HDU 4254 | A Famous Game |
|
31 / 141 | Problem J | HDU 4268 | Alice and Bob |
|
23 / 46 | Problem K | HDU 4260 | The End of The World |
|
56 / 167 | Problem L | CodeForces 557B | Pasha and Tea |
A
题意：有C类课程，每类课程有T个班，不同的班在不同位置，上不同的班有不同的花费，初始位置是0，末位置L，求最小花费。
比较简单的递推，f[i][j]表示上i类课的j个班所需的最小花费。最后求f[c]中的最小值就可以了。
1 #include <iostream> 2 #include <cstdio> 3 #include <fstream> 4 #include <algorithm> 5 #include <cmath> 6 #include <deque> 7 #include <vector> 8 #include <queue> 9 #include <string> 10 #include <cstring> 11 #include <map> 12 #include <stack> 13 #include <set> 14 #define LL long long 15 #define eps 1e-8 16 #define INF 0x3f3f3f3f 17 //#define OPEN_FILE 18 using namespace std; 19 int T, c, t, L; 20 struct Node{ 21 int pos, v; 22 }p[30][1005]; 23 int f[30][1005]; 24 int main() 25 { 26 #ifdef OPEN_FILE 27 freopen("in.txt", "r", stdin); 28 //freopen("out.txt", "w", stdout); 29 #endif // OPEN_FILE 30 scanf("%d", &T); 31 for (int cas = 1; cas <= T; cas++){ 32 scanf("%d%d%d", &c, &t, &L); 33 for (int i = 1; i <= c; i++){ 34 for (int j = 1; j <= t; j++){ 35 scanf("%d%d", &p[i][j].pos, &p[i][j].v); 36 } 37 } 38 memset(f, INF, sizeof(f)); 39 for (int i = 1; i <= t; i++){ 40 f[1][i] = p[1][i].pos + p[1][i].v; 41 } 42 for (int i = 2; i <= c; i++){ 43 for (int j = 1; j <= t; j++){ 44 for (int k = 1; k <= t; k++){ 45 int dis = abs(p[i - 1][k].pos - p[i][j].pos); 46 f[i][j] = min(f[i][j], f[i - 1][k] + dis + p[i][j].v); 47 } 48 } 49 } 50 int ans = INF; 51 for (int i = 1; i <= t; i++){ 52 int dis = abs(p[c][i].pos - L); 53 ans = min(ans, f[c][i] + dis); 54 } 55 printf("%d ", ans); 56 } 57 }
B
题意:给你一个长度为n的数列，循环起来相当于n个数列，求这n个数列里所有前缀和都非负的有多少
单调队列。
1 #include <iostream> 2 #include <cstdio> 3 #include <fstream> 4 #include <algorithm> 5 #include <cmath> 6 #include <deque> 7 #include <vector> 8 #include <queue> 9 #include <string> 10 #include <cstring> 11 #include <map> 12 #include <stack> 13 #include <set> 14 #define LL long long 15 #define eps 1e-8 16 #define INF 0x3f3f3f3f 17 //#define OPEN_FILE 18 #define MAXN 2000005 19 using namespace std; 20 int n; 21 int a[MAXN], Q[MAXN]; 22 LL sum[MAXN]; 23 int main() 24 { 25 #ifdef OPEN_FILE 26 freopen("in.txt", "r", stdin); 27 //freopen("out.txt", "w", stdout); 28 #endif // OPEN_FILE 29 while (~scanf("%d", &n)){ 30 if (n == 0) break; 31 memset(a, 0, sizeof(a)); 32 memset(sum, 0, sizeof(sum)); 33 for (int i = 1; i <= n; i++){ 34 scanf("%d", &a[i]); 35 a[i + n] = a[i]; 36 } 37 for (int i = 1; i <= 2 * n; i++){ 38 sum[i] = sum[i - 1] + a[i]; 39 } 40 int head = 0, tail = 0; 41 int ans = 0; 42 for (int i = 1; i <= 2 * n; i++){ 43 //keep queue's order 44 while (head < tail && sum[Q[tail - 1]] >= sum[i]){ 45 tail--; 46 } 47 Q[tail++] = i; 48 if (i > n && sum[Q[head]] - sum[i - n] >= 0) { 49 ans++; 50 } 51 //keep n numbers in queue 52 while (head < tail && i - Q[head] >= n){ 53 head++; 54 } 55 } 56 printf("%d ", ans); 57 } 58 }
C
题意：coming soon
D
题意：给你一个坐标表示你的初始位置，再给你一些直线，问沿着这些线最远能离初始位置多远。
E
题意：给你一棵树，边的权值在[0,L]中随机，求使树的总权值不超过S的概率
F
题意：coming soon
G
题意：给你n表示置换循环的长度，K表示1在这个循环置换中的对应最大值，问形成这种置换的排列数。
H
题意：给你一棵树，边由两个公司组成，问使的正好有K条A公司的边的最小生成树的花费。
I
题意：袋子里有N个球，每个球是蓝色或者是红色，取出来P个球里面有Q个红球，问再拿出一个球来是红球的概率
J
题意：A有N张牌，B有M张牌，一张牌能盖住另一张牌的条件是height和width都不比另一张牌小，为A最多能盖住B多少张牌。
K
题意：从小到大给你汉诺塔中盘子位于哪个柱子，问全部移到B柱上要多少步。
work（a,b,t)表示把第t个盘从a移到b，如果t-1号盘在a，那就只要把这t-1个盘移到另外一个盘，再把t移到b，再把t-1个放好的盘移到b，后面的这个过程是标准的汉娜塔就直接可以得到步数
1 #include <iostream> 2 #include <cstdio> 3 #include <fstream> 4 #include <algorithm> 5 #include <cmath> 6 #include <deque> 7 #include <vector> 8 #include <queue> 9 #include <string> 10 #include <cstring> 11 #include <map> 12 #include <stack> 13 #include <set> 14 #define LL long long 15 #define eps 1e-8 16 #define INF 0x3f3f3f3f 17 #define OPEN_FILE 18 using namespace std; 19 const char CH[3] = { 'A', 'B', 'C' }; 20 LL ans; 21 char s[70]; 22 LL POW(int p){ 23 LL res = 1; 24 for (int i = 1; i <= p; i++){ 25 res *= 2; 26 } 27 return res; 28 } 29 void work(char a, char b, LL m){ 30 if (m == 0){ 31 return; 32 } 33 int i; 34 for (i = 0; i < 3; i++){ 35 if (a != CH[i] && b != CH[i]) break; 36 } 37 if (s[m - 1] == a){ 38 ans += POW(m - 1); 39 work(a, CH[i], m - 1); 40 } 41 else if (s[m - 1] == b){ 42 work(CH[i], b, m - 1); 43 } 44 else{ 45 ans += POW(m - 1); 46 work(CH[i], a, m - 1); 47 } 48 } 49 50 int main() 51 { 52 #ifdef OPEN_FILE 53 //freopen("in.txt", "r", stdin); 54 //freopen("out.txt", "w", stdout); 55 #endif // OPEN_FILE 56 while (~scanf("%s", s)){ 57 if (s[0] == 'X') break; 58 ans = 0; 59 work('A', 'B', strlen(s)); 60 printf("%I64d ", ans); 61 } 62 63 }
L
题意：有M升水到倒给2N个人，N个男生，N个女生，男生杯子里的水必须是女生的两倍，男生之间和女生之间的水必须一样，问最多倒出多少升。
比较简单的贪心。
1 #include <iostream> 2 #include <cstdio> 3 #include <fstream> 4 #include <algorithm> 5 #include <cmath> 6 #include <deque> 7 #include <vector> 8 #include <queue> 9 #include <string> 10 #include <cstring> 11 #include <map> 12 #include <stack> 13 #include <set> 14 #define LL long long 15 #define eps 1e-8 16 #define INF 0x3f3f3f3f 17 #define OPEN_FILE 18 using namespace std; 19 int n; 20 double w; 21 double a[200005]; 22 int main() 23 { 24 #ifdef OPEN_FILE 25 //freopen("in.txt", "r", stdin); 26 // freopen("out.txt", "w", stdout); 27 #endif // OPEN_FILE 28 scanf("%d%lf", &n, &w); 29 for (int i = 1; i <= 2 * n; i++){ 30 scanf("%lf", &a[i]); 31 } 32 sort(a + 1, a + 2 * n + 1); 33 double ans = 0; 34 if (a[n + 1] > a[1] * 2){ 35 if (a[1] * n * 3 <= w){ 36 ans = a[1] * n * 3; 37 } 38 else{ 39 ans = w; 40 } 41 } 42 else{ 43 if (a[n + 1] * n * 1.5 <= w){ 44 ans = a[n + 1] * n * 1.5; 45 } 46 else{ 47 ans = w; 48 } 49 } 50 printf("%.6lf ", ans); 51 52 }