Google KickStart 2020 RoundA
第一题：Allocation
题意：N个房子，第i个房子价格为Ai美元，用B美元最多买几个房子。
Limit：
1 ≤ T ≤ 100.
1 ≤ B ≤ 10^5.
1 ≤ Ai ≤ 1000, for all i.
1 ≤ N ≤ 10^5.
分析：按价格从小到大排序即可。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 100000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a < b ? -1 : 1;
}
int a[MAXN];
int main(){
int T;
scanf("%d", &T);
for(int Case = 1; Case <= T; ++Case){
int N, B;
scanf("%d%d", &N, &B);
for(int i = 0; i < N; ++i){
scanf("%d", &a[i]);
}
sort(a, a + N);
int sum = 0;
int cnt = 0;
for(int i = 0; i < N; ++i){
if(sum + a[i] <= B){
sum += a[i];
++cnt;
}
else break;
}
printf("Case #%d: %d\n", Case, cnt);
}
return 0;
}
第二题：Plates
题意：有N堆盘子，每堆k个盘子，每个盘子有个beauty value。当拿某个盘子时，该盘子之上的盘子也会被拿起。问当拿P个盘子时，能获得最大的beauty value值。
Limit：
1 ≤ T ≤ 100.
1 ≤ K ≤ 30.
1 ≤ P ≤ N * K.
The beauty values are between 1 and 100, inclusive.
1 ≤ N ≤ 50.
分析：dp[i][j]表示前i堆拿j个盘子能获得的最大beauty value值。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 100000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a < b ? -1 : 1;
}
int a[55][35];
int dp[55][1510]; //dp数组第二维比赛时开小了
int sum[55][35];
int main(){
int T;
scanf("%d", &T);
for(int Case = 1; Case <= T; ++Case){
int N, K, P;
scanf("%d%d%d", &N, &K, &P);
memset(sum, 0, sizeof sum);
for(int i = 1; i <= N; ++i){
for(int j = 1; j <= K; ++j){
scanf("%d", &a[i][j]);
sum[i][j] = sum[i][j - 1] + a[i][j];
}
}
memset(dp, 0, sizeof dp);
for(int i = 1; i <= N; ++i){
for(int j = 0; j <= P; ++j){
dp[i][j] = max(dp[i][j], dp[i - 1][j]);
for(int k = 1; k <= K && k <= j; ++k){
dp[i][j] = max(dp[i][j], dp[i - 1][j - k] + sum[i][k]);
}
}
}
printf("Case #%d: %d\n", Case, dp[N][P]);
}
return 0;
}
优化一下，只开一维dp数组~
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 100000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a < b ? -1 : 1;
}
int a[55][35];
int dp[1510];
int sum[55][35];
int main(){
int T;
scanf("%d", &T);
for(int Case = 1; Case <= T; ++Case){
int N, K, P;
scanf("%d%d%d", &N, &K, &P);
memset(sum, 0, sizeof sum);
for(int i = 1; i <= N; ++i){
for(int j = 1; j <= K; ++j){
scanf("%d", &a[i][j]);
sum[i][j] = sum[i][j - 1] + a[i][j];
}
}
memset(dp, 0, sizeof dp);
for(int i = 1; i <= N; ++i){
for(int j = P; j >= 0; --j){
for(int k = 1; k <= K && k <= j; ++k){
dp[j] = max(dp[j], dp[j - k] + sum[i][k]);
}
}
}
printf("Case #%d: %d\n", Case, dp[P]);
}
return 0;
}
第三题：Workout
题意：有N个session，每个session长达Mi分钟。定义diff为相邻session差值的最大值，要求增加至多K个session，使得diff最小，求最小的diff。（N个session的时长是严格递增的，要求增加k个session后，所有session时长也是严格递增的）
Limit：
1 ≤ T ≤ 100.
For at most 10 test cases, 2 ≤ N ≤ 10^5.
For all other test cases, 2 ≤ N ≤ 300.
1 ≤ Mi ≤ 10^9.
Mi < Mi+1 for all i.
1 ≤ K ≤ 10^5.
分析：二分diff求右临界值即可。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 100000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a < b ? -1 : 1;
}
int a[MAXN];
int N, K;
bool judge(int x){
int cnt = 0;
for(int i = 2; i <= N; ++i){
if(a[i] - a[i - 1] > x){
int tmp = a[i - 1];
while(tmp + x < a[i]){
++cnt;
if(cnt > K) return false;
tmp += x;
}
}
}
return true;
}
int main(){
int T;
scanf("%d", &T);
for(int Case = 1; Case <= T; ++Case){
scanf("%d%d", &N, &K);
int diff = 0;
for(int i = 1; i <= N; ++i){
scanf("%d", &a[i]);
if(i > 1){
diff = max(diff, a[i] - a[i - 1]);
}
}
int l = 1;
int r = diff;
while(l < r){
int mid = l + (r - l) / 2;
if(judge(mid)) r = mid;
else l = mid + 1;
}
printf("Case #%d: %d\n", Case, r);
}
return 0;
}
第四题：Bundling
题意：有N个字符串，每个字符串只包含字母A~Z，要求把字符串按K个一组进行分组，每个字符串只能分到一个组中。分好后，每个组的score为该组所有字符串的最长公共前缀。问分组后每个组的score之和的最大值。
Limit：
1 ≤ T ≤ 100.
2 ≤ N ≤ 105.
2 ≤ K ≤ N.
K divides N.
Each of Pip's strings contain at least one character.
Each string consists only of letters from A to Z.
The total number of characters in Pip's strings across all test cases is at most 2 × 106.
分析：
（1）对于N个字符串中所有可能的前缀来说，假如某个前缀p存在于cnt个字符串中，则它对最终答案的贡献是cnt/K。累加所有的前缀贡献即可。
（2）计算每个前缀存在于多少个字符串中，用Trie树即可。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 100000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a < b ? -1 : 1;
}
struct TreeNode{
int val;
TreeNode *nex[26];
TreeNode(){
val = 0;
for(int i = 0; i < 26; ++i) nex[i] = NULL;
}
};
string s[MAXN];
void build(string x, TreeNode* root){
int len = x.size();
for(int i = 0; i < len; ++i){
int cur = x[i] - 'A';
if(root -> nex[cur] == NULL){
root -> nex[cur] = new TreeNode();
}
root = root -> nex[cur];
++root -> val;
}
}
int bfs(TreeNode* root, int K){
int ans = 0;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()){
TreeNode *cur = q.front();
q.pop();
ans += cur -> val / K;
for(int i = 0; i < 26; ++i){
if(cur -> nex[i] != NULL){
q.push(cur -> nex[i]);
}
}
}
return ans;
}
int main(){
int T;
scanf("%d", &T);
for(int Case = 1; Case <= T; ++Case){
int N, K;
scanf("%d%d", &N, &K);
TreeNode *root = new TreeNode();
for(int i = 0; i < N; ++i){
cin >> s[i];
build(s[i], root);
}
printf("Case #%d: %d\n", Case, bfs(root, K));
}
return 0;
}
也可以省掉bfs~
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
typedef long long LL;
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 100000 + 10;
const double eps = 1e-8;
int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a < b ? -1 : 1;
}
struct TreeNode{
int val;
TreeNode *nex[26];
TreeNode(){
val = 0;
for(int i = 0; i < 26; ++i) nex[i] = NULL;
}
};
string s[MAXN];
void build(string x, TreeNode* root, int K, int &ans){
int len = x.size();
for(int i = 0; i < len; ++i){
int cur = x[i] - 'A';
if(root -> nex[cur] == NULL){
root -> nex[cur] = new TreeNode();
}
root = root -> nex[cur];
++root -> val;
if(root -> val == K){
++ans;
root -> val = 0;
}
}
}
int main(){
int T;
scanf("%d", &T);
for(int Case = 1; Case <= T; ++Case){
int N, K;
scanf("%d%d", &N, &K);
TreeNode *root = new TreeNode();
int ans = 0;
for(int i = 0; i < N; ++i){
cin >> s[i];
build(s[i], root, K, ans);
}
printf("Case #%d: %d\n", Case, ans);
}
return 0;
}
