牛客算法周周练2
牛客算法周周练2
B Music Problem
题意：n个物品 01选法 能否凑成3600的整数倍。
思路：dp（写麻烦了，不需要二进制优化其实也可以过，因为抽屉原理当n>=3600时肯定存在答案）
很简单的一个dp，，，但是学到了bitset这种牛逼偷懒 的写法。
dp[i] 表示当前容量是否可行 即dp[j]=(backup[j]|backup[((j-a[i])%3600+3600)%3600]);
dp代码
#pragma GCC optimize(3,"Ofast","inline") //G++
#include
using namespace std;
//======================================
namespace FastIO {char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh1 = '\n', hh2 = ' ';int p, p3 = -1;
void read() {}
void print() {buf2[p3] = hh1;}
inline int getc() {return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;}
inline void flush() {fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;}
template <typename T, typename... T2>
inline void read(T &x, T2 &... oth) {int f = 0;x = 0;char ch = getc();while (!isdigit(ch)) {if (ch == '-')f = 1;ch = getc();}while (isdigit(ch)) {x = x * 10 + ch - 48;ch = getc();}x = f ? -x : x;read(oth...);}
template <typename T, typename... T2>
inline void print(T x, T2... oth) {if (p3 > 1 << 20)flush();if (x < 0)buf2[++p3] = 45, x = -x;do {a[++p] = x % 10 + 48;} while (x /= 10);do {buf2[++p3] = a[p];} while (--p);buf2[++p3] = hh2;print(oth...);}} // namespace FastIO
using FastIO::print;
using FastIO::read;
typedef long long ll;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn = 1e6+5;
//======================================
int mp[maxn];
int dp[10005],backup[10005];
int a[maxn];
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
clock_t c1 = clock();
#endif
//**************************************
int T;
read(T);
while(T--){
int n;
read(n);
for(int i=0,x;i<n;i++){
read(x);
mp[x%3600]++;
}
mp[3600]=mp[0];
int cnt=0;
for(int i=1;i<=3600;i++){
if(!mp[i]) continue;
int k=1;
while(k<=mp[i]){
a[++cnt]=i*k;
mp[i]-=k,k*=2;
}
if(mp[i]){
a[++cnt]=mp[i]*i;
mp[i]=0;
}
}
n=cnt;
backup[0]=1;
for(int i=1;i<=n;i++){
for(int j=0;j<=3600;j++){
dp[j]=(backup[j]|backup[((j-a[i])%3600+3600)%3600]);
}
memcpy(backup,dp,sizeof dp);
if(dp[3600]) break;
}
puts((dp[3600])?"YES":"NO");
for(int i=0;i<=3600;i++) mp[i]=0,dp[i]=0,backup[i]=0;
}
//**************************************
FastIO::flush();
#ifndef ONLINE_JUDGE
cerr << "Time:" << clock() - c1 << "ms" << endl;
#endif
return 0;
}
bitset代码
#pragma GCC optimize(3,"Ofast","inline") //G++
#include
using namespace std;
//======================================
namespace FastIO {char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh1 = '\n', hh2 = ' ';int p, p3 = -1;
void read() {}
void print() {buf2[p3] = hh1;}
inline int getc() {return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;}
inline void flush() {fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;}
template <typename T, typename... T2>
inline void read(T &x, T2 &... oth) {int f = 0;x = 0;char ch = getc();while (!isdigit(ch)) {if (ch == '-')f = 1;ch = getc();}while (isdigit(ch)) {x = x * 10 + ch - 48;ch = getc();}x = f ? -x : x;read(oth...);}
template <typename T, typename... T2>
inline void print(T x, T2... oth) {if (p3 > 1 << 20)flush();if (x < 0)buf2[++p3] = 45, x = -x;do {a[++p] = x % 10 + 48;} while (x /= 10);do {buf2[++p3] = a[p];} while (--p);buf2[++p3] = hh2;print(oth...);}} // namespace FastIO
using FastIO::print;
using FastIO::read;
typedef long long ll;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn = 1e6+5;
//======================================
int mp[maxn];
int a[maxn];
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
clock_t c1 = clock();
#endif
//**************************************
int T;
read(T);
while(T--){
int n;
read(n);
for(int i=0,x;i<n;i++){
read(x);
mp[x%3600]++;
}
int cnt=0;
for(int i=0;i<3600;i++){
if(!mp[i]) continue;
int k=1;
while(k<=mp[i]){
a[++cnt]=i*k;
mp[i]-=k,k*=2;
}
if(mp[i]){
a[++cnt]=mp[i]*i;
mp[i]=0;
}
}
n=cnt;
bitset<3600>res,cur;
for(int i=1;i<=n;i++){
int x=a[i]%3600;
cur.reset(),cur[x]=1;
cur|=(res<<x)|(res>>(3600-x));
res|=cur;
if(res[0]) break;
}
puts(res[0]?"YES":"NO");
}
//**************************************
FastIO::flush();
#ifndef ONLINE_JUDGE
cerr << "Time:" << clock() - c1 << "ms" << endl;
#endif
return 0;
}
题意：树形结构，q次操作。每次增加u和与u距离小于等于2的点权，增加1。在线询问u此时的点权
思路：维护dp数组 ,dp[x][0]表示x对自己的贡献，dp[x][1]表示x对与自已距离为1的儿子的贡献，dp[x][2]表示x对与自已距离为2的孙子的贡献
修改为： dp[fa[x]][1]++; dp[fa[x]][0]++; dp[fa[fa[x]]][0]++; dp[x][1]++; dp[x][2]++;
查询为：dp[x][0]+dp[fa[x]][1]+dp[fa[fa[x]]][2]
#pragma GCC optimize(3,"Ofast","inline") //G++
#include
using namespace std;
//======================================
namespace FastIO {char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh1 = '\n', hh2 = ' ';int p, p3 = -1;
void read() {}
void print() {buf2[p3] = hh1;}
inline int getc() {return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;}
inline void flush() {fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;}
template <typename T, typename... T2>
inline void read(T &x, T2 &... oth) {int f = 0;x = 0;char ch = getc();while (!isdigit(ch)) {if (ch == '-')f = 1;ch = getc();}while (isdigit(ch)) {x = x * 10 + ch - 48;ch = getc();}x = f ? -x : x;read(oth...);}
template <typename T, typename... T2>
inline void print(T x, T2... oth) {if (p3 > 1 << 20)flush();if (x < 0)buf2[++p3] = 45, x = -x;do {a[++p] = x % 10 + 48;} while (x /= 10);do {buf2[++p3] = a[p];} while (--p);buf2[++p3] = hh2;print(oth...);}} // namespace FastIO
using FastIO::print;
using FastIO::read;
typedef long long ll;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn = 1e6+5;
//======================================
vector<int>edge[maxn];
int fa[maxn];
int dp[maxn][3];
void addedge(int x,int y){
edge[x].push_back(y);
}
void dfs(int u,int father){
fa[u]=father;
for(auto v:edge[u]){
if(v==father) continue;
dfs(v,u);
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
clock_t c1 = clock();
#endif
//**************************************
int n,q;
read(n,q);
for(int i=1,x,y;i<=n-1;i++){
read(x,y);
addedge(x,y),addedge(y,x);
}
dfs(1,0);
while(q--){
int x;
read(x);
dp[fa[x]][1]++;
dp[fa[x]][0]++;
dp[fa[fa[x]]][0]++;
dp[x][1]++;
dp[x][2]++;
print(dp[x][0]+dp[fa[x]][1]+dp[fa[fa[x]]][2]);
}
//**************************************
FastIO::flush();
#ifndef ONLINE_JUDGE
cerr << "Time:" << clock() - c1 << "ms" << endl;
#endif
return 0;
}