yxf 的 信息学奥赛复赛练习2
hanoi双塔问题
思路
递推鸟题
\[F_0=0\\
F_1=1\\
F_n=2F_{n-2}+2
\]
时间复杂度
预处理 \(O(n)\)
作答 \(O(1)\)
蒟蒻代码
#include <bits/stdc++.h>
#define re register
using namespace std;
typedef long long ll;
const int N=410;
int n;
ll ans[N];
void make_table(){
ans[0]=0; ans[1]=1;
for(re int i=1;i<=405;i++)
ans[i]=2*ans[i-2]+2;
}
int main()
{
ios::sync_with_stdio(0);
clock_t c1 = clock();
#ifdef LOCAL
freopen("data.in","r",stdin);
freopen("data.out","w",stdout);
#endif
// ======================================================================
make_table();
cin>>n;
cout<<ans[n<<1];
// ======================================================================
end:
cerr << "Time Used:" << clock() - c1 << "ms" << endl;
return 0;
}
天使的起誓
思路
m 每位处理，利用 mod 运算性质
时间复杂度
\(O(\log_{2}m)\)
蒟蒻代码
#include <bits/stdc++.h>
#define re register
using namespace std;
const int N=1e3+10;
int n;
int ans=0;
string m;
int base=1;
int main()
{
ios::sync_with_stdio(0);
clock_t c1 = clock();
#ifdef LOCAL
freopen("data.in","r",stdin);
freopen("data.out","w",stdout);
#endif
// ======================================================================
cin>>n>>m;
for(re int i=m.length()-1;i>=0;i--){
ans=(ans+(m[i]-'0')*base)%n;
base=base*10%n;
}
if(!ans) cout<<n;
else cout<<ans;
// ======================================================================
end:
cerr << "Time Used:" << clock() - c1 << "ms" << endl;
return 0;
}
飞跃原野
思路
三维 BFS 鸟题
蒟蒻代码
#include <bits/stdc++.h>
#define re register
#define LOCAL
using namespace std;
struct rec{
int x;
int y;
int cnt;
int rest; // 剩下飞行
rec(int _x,int _y,int _cnt,int _rest):x(_x),y(_y),cnt(_cnt),rest(_rest){}
};
const int N=110;
const int M=110;
const int D=110;
const int dx[4]={-1,0,0,1};
const int dy[4]={0,-1,1,0};
char a[N][M];
int n,m,d;
bool f[N][M][D]; // 到 (m,n) 位置且 还剩下 d 次飞行是否搜索过
queue<rec> q;
inline bool valid(int x,int y){
return x>0&&x<=m&&y>0&&y<=n&&a[y][x]=='P';
}
int main(){
ios::sync_with_stdio(0);
# ifdef LOCAL
freopen("fly.in","r",stdin);
freopen("fly.out","w",stdout);
# endif
// ====================================================
cin>>n>>m>>d;
for(re int i=1;i<=n;i++)
for(re int j=1;j<=m;j++)
cin>>a[i][j];
memset(f,0,sizeof(f));
q.push(rec(1,1,0,d));
while(!q.empty()){
rec r=q.front(); q.pop();
if(r.x==m&&r.y==n){
cout<<r.cnt<<endl;
goto end;
}
for(re int i=0;i<4;i++){
int nx,ny;
// 走路
nx=r.x+dx[i];
ny=r.y+dy[i];
if(valid(nx,ny) && !f[ny][nx][r.rest]){
f[ny][nx][r.rest]=1;
q.push(rec(nx,ny,r.cnt+1,r.rest));
}
// 飞行
for(re int j=2;j<=r.rest;j++){
nx=r.x+j*dx[i];
ny=r.y+j*dy[i];
if(valid(nx,ny) && !f[ny][nx][r.rest-j]){
f[ny][nx][r.rest-j]=1;
q.push(rec(nx,ny,r.cnt+1,r.rest-j));
}
}
}
}
puts("impossible");
// ====================================================
end:
return 0;
}
yxf 的 \(20\) 分代码
依旧, 不出所料, yxf写的 BFS 又让他抓耳挠腮了......
找半天错没找出来......
此处挂上 yxf 的错误代码
雾
#include<bits/stdc++.h>
using namespace std;
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int a[105][105];
int q[1000005][5];
bool v[105][105][105];
int n,m,k;
int bfs(int x,int y,int s,int r)
{
int front=0,rear=0;
int xx,yy;
rear++;
q[rear][1]=x;
q[rear][2]=y;
q[rear][3]=s;
q[rear][4]=r;
v[1][1][r]=true;
while(front<rear)
{
front++;
x=q[front][1];
y=q[front][2];
r=q[front][4];
for(int i=0;i<=3;i++)
{
xx=x+dx[i];
yy=y+dy[i];
if((a[xx][yy]==1)&&(!v[xx][yy][r]))
{
v[xx][yy][r]=true;
rear++;
q[rear][1]==xx;
q[rear][2]=yy;
q[rear][3]=q[front][3]+1;
q[rear][4]=r;
if(xx==n&&yy==m)
{
cout<<q[rear][3];
return 0;
}
}
for(int j=2;j<=r;j++)
{
xx=x+dx[i]*j;
yy=y+dy[i]*j;
if(a[xx][yy]==0)break;
if(a[xx][yy]==1&&!v[xx][yy][r-j])
{
v[xx][yy][r-j]=true;
rear++;
q[rear][1]==xx;
q[rear][2]=yy;
q[rear][3]=q[front][3]+1;
q[rear][4]=r-j;
if(xx==n&&yy==m)
{
cout<<q[rear][3];
return 0;
}
}
}
}
}
cout<<"impossible";
}
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
cin>>n>>m>>k;
char ch;
scanf("%c",&ch);//读掉回车
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++) {
scanf("%c",&ch);
if(ch=='P')a[i][j]=1;else a[i][j]=0;
}
scanf("%c",&ch);
}
if(a[1][1]!=1)
{
cout<<"impossible";
}
else
{
bfs(1,1,0,k);
}
return 0;
}
关系网络
思路
就建图，最后输出 \(SSSP(x\rightarrow y)-1\)
蒟蒻代码
#include <bits/stdc++.h>
#define re register
using namespace std;
struct Edge{
int to;
int nxt;
};
struct node{
int ver;
int d;
node(int Ver,int D):ver(Ver),d(D){}
bool operator <(const node& n1) const{
return d>n1.d;
}
};
const int N=110;
int n,x,y;
int tot=0,head[N];
Edge edge[N*N];
int dis[N];
bitset<N> vis;
priority_queue<node> pq;
void add(int x,int y){
edge[++tot].to=y;
edge[tot].nxt=head[x];
head[x]=tot;
}
int main()
{
ios::sync_with_stdio(0);
clock_t c1 = clock();
#ifdef LOCAL
freopen("data.in","r",stdin);
freopen("data.out","w",stdout);
#endif
// ======================================================================
cin>>n>>x>>y;
memset(dis,0x3f,sizeof(dis));
for(re int i=1;i<=n;i++){
for(re int j=1;j<=n;j++){
int tmp; cin>>tmp;
if(tmp) add(i,j);
}
}
// x->y
dis[x]=0; pq.push(node(x,0));
while(!pq.empty()){
int a=pq.top().ver; pq.pop();
if(vis[a]) continue;
vis[a]=1;
for(re int i=head[a];i;i=edge[i].nxt){
int y=edge[i].to;
if(dis[a]+1<dis[y]){
dis[y]=dis[a]+1;
pq.push(node(y,dis[y]));
}
}
}
cout<<dis[y]-1;
// ======================================================================
end:
cerr << "Time Used:" << clock() - c1 << "ms" << endl;
return 0;
}
