2015 UESTC Winter Training 中的简单搜索
uvalive 4997
简答搜索
先将中心求出来，并且染上无关色，染色之后若有地方未染色，则说明不行
然后再用回溯法染色
// whn6325689
// Mr.Phoebe
// http://blog.csdn.net/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;
#define CLR(x,y) memset(x,y,sizeof(x))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))
#define MID(x,y) (x+((y-x)>>1))
#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
template<class T>
inline bool read(T &n)
{
T x = 0, tmp = 1;
char c = getchar();
while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
if(c == EOF) return false;
if(c == '-') c = getchar(), tmp = -1;
while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
n = x*tmp;
return true;
}
template <class T>
inline void write(T n)
{
if(n < 0)
{
putchar('-');
n = -n;
}
int len = 0,data[20];
while(n)
{
data[len++] = n%10;
n /= 10;
}
if(!len) data[len++] = 0;
while(len--) putchar(data[len]+48);
}
//-----------------------------------
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
const int DX[] = {-2, -2, -2, -1, -1, -1, -1, 0, 0, 1, 1, 1, 1, 2, 2, 2};
const int DY[] = {-1, 0, 1, -2, -1, 1, 2, -2, 2, -2, -1, 1, 2, -1, 0, 1};
int n;
char ma[20][20];
struct Node
{
int x, y;
char col;
Node(){}
Node(const int &x,const int &y,const char col='\0'):x(x),y(y),col(col){}
} a[400];
int tot;
bool flag;
bool isCenter(int x, int y)
{
if(ma[x][y] != '.') return 0;
for(int k=0; k<4; k++)
{
int xx=x+dx[k];
int yy=y+dy[k];
if(xx<1 || xx>n || yy<1 || yy>n)
return 0;
if(ma[xx][yy]!='.')
return 0;
}
return 1;
}
void change(int x,int y,char ch)
{
ma[x][y]=ch;
for(int k=0; k<4; k++)
ma[x+dx[k]][y+dy[k]]=ch;
}
bool isfull()
{
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
if(ma[i][j]=='.')
return 0;
return 1;
}
bool check(int x,int y,char ch)
{
for(int k=0; k<16; k++)
{
int xx=x+DX[k];
int yy=y+DY[k];
if(xx<1 || xx>n || yy<1 || yy>n) continue;
if(ma[xx][yy]==ch) return 0;
}
return 1;
}
void dfs(int x)
{
if(flag) return;
if(x>tot)
{
flag=1;
return;
}
for(char ch='B'; ch<='D'; ch++)
{
if(!check(a[x].x,a[x].y,ch)) continue;
a[x].col=ch;
change(a[x].x,a[x].y,ch);
dfs(x+1);
if(flag)
return;
}
change(a[x].x,a[x].y,'q');
}
int main()
{
int T,cas=1;
read(T);
while(T--)
{
read(n);
printf("Case %d:",cas++);
for(int i=1; i<=n; i++)
scanf("%s", ma[i]+1);
tot=0;
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
if(isCenter(i, j))
{
a[++tot]=Node(i,j);
change(i,j,'q');
}
if(!isfull())
puts(" Not Possible!");
else
{
flag=0;
dfs(1);
if(!flag)
puts(" Not Possible!");
else
{
putchar('\n');
for(int i=1; i<=n; i++)
printf("%s\n",ma[i]+1);
}
}
}
return 0;
}
bfs求最短路
然后用背包求答案
// whn6325689
// Mr.Phoebe
// http://blog.csdn.net/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;
#define CLR(x,y) memset(x,y,sizeof(x))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))
#define MID(x,y) (x+((y-x)>>1))
#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
template<class T>
inline bool read(T &n)
{
T x = 0, tmp = 1; char c = getchar();
while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
if(c == EOF) return false;
if(c == '-') c = getchar(), tmp = -1;
while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
n = x*tmp;
return true;
}
template <class T>
inline void write(T n)
{
if(n < 0)
{
putchar('-');
n = -n;
}
int len = 0,data[20];
while(n)
{
data[len++] = n%10;
n /= 10;
}
if(!len) data[len++] = 0;
while(len--) putchar(data[len]+48);
}
//-----------------------------------
struct Node
{
int x,y,z;
int step;
Node(const int& x=0,const int& y=0,const int& z=0,const int& step=0):x(x),y(y),z(z),step(step){}
};
struct P
{
int ti,sc;
}pp[111];
const int dx[]={0,0,0,0,1,-1},dy[]={1,-1,0,0,0,0},dz[]={0,0,1,-1,0,0};
char ma[15][111][111];
bool vis[15][111][111];
int dp[10010],p[15][111][111];
int l,h,w,n,s,tot;
Node st;
queue<Node> q;
bool in(int x,int y,int z)
{
if (x>=1 && x<=l && y>=1 && y<=h && z>=1 && z<=w && ma[x][y][z]!='X' && vis[x][y][z]==0)
return 1;
return 0;
}
void bfs()
{
int x,y,z;
CLR(vis,0);
while(!q.empty()) q.pop();
q.push(st);
vis[st.x][st.y][st.z]=1;
tot=0;
while(!q.empty())
{
Node now=q.front();
q.pop();
if(ma[now.x][now.y][now.z]=='#')
{
pp[++tot].sc=p[now.x][now.y][now.z];
pp[tot].ti=now.step*3;
}
for(int i=0;i<4;i++)
{
x=now.x+dx[i];
y=now.y+dy[i];
z=now.z+dz[i];
if(in(x,y,z))
{
q.push(Node(x,y,z,now.step+1));
vis[x][y][z]=1;
}
}
if(ma[now.x][now.y][now.z]=='U')
{
x=now.x+dx[4];
y=now.y+dy[4];
z=now.z+dz[4];
if(in(x,y,z))
{
q.push(Node(x,y,z,now.step+1));
vis[x][y][z]=1;
}
}
if(ma[now.x][now.y][now.z]=='D')
{
x=now.x+dx[5];
y=now.y+dy[5];
z=now.z+dz[5];
if(in(x,y,z))
{
q.push(Node(x,y,z,now.step+1));
vis[x][y][z]=1;
}
}
}
}
int main()
{
// freopen("data.txt","r",stdin);
int T;
scanf("%d",&T);
while(T--)
{
CLR(dp,0);CLR(pp,0);
scanf("%d %d %d %d %d",&l,&h,&w,&n,&s);
for(int i=1;i<=l;i++)
for(int j=1;j<=h;j++)
{
scanf("%s",ma[i][j]+1);
for(int k=1;k<=w;k++)
if(ma[i][j][k]=='S')
st=Node(i,j,k,0);
}
for(int i=1,fl,hh,ww,sc;i<=n;i++)
{
scanf("%d %d %d %d",&fl,&hh,&ww,&sc);
p[fl][hh][ww]=sc;
ma[fl][hh][ww]='#';
}
/*
for(int i=1;i<=l;i++)
for(int j=1;j<=h;j++)
printf("%s\n",ma[i][j]+1);
*/
bfs();
for(int i=1;i<=tot;i++)
for(int j=s;j>=pp[i].ti;j--)
dp[j]=max(dp[j],dp[j-pp[i].ti]+pp[i].sc);
printf("%d\n",dp[s]);
}
return 0;
}
HDU 4166
bfs暴搜
本来想用A*的，后来想起来还要求路径
// whn6325689
// Mr.Phoebe
// http://blog.csdn.net/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;
#define CLR(x,y) memset(x,y,sizeof(x))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))
#define MID(x,y) (x+((y-x)>>1))
#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
template<class T>
inline bool read(T &n)
{
T x = 0, tmp = 1; char c = getchar();
while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
if(c == EOF) return false;
if(c == '-') c = getchar(), tmp = -1;
while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
n = x*tmp;
return true;
}
template <class T>
inline void write(T n)
{
if(n < 0)
{
putchar('-');
n = -n;
}
int len = 0,data[20];
while(n)
{
data[len++] = n%10;
n /= 10;
}
if(!len) data[len++] = 0;
while(len--) putchar(data[len]+48);
}
//-----------------------------------
int n,m,mod;
const int dir[4][2]={-1,0,0,1,1,0,0,-1};
struct Node
{
int x,y,dir;
int step;
Node(const int& x=0,const int& y=0,const int& dir=0,const int& step=0):x(x),y(y),dir(dir),step(step){}
bool operator == (const Node& b) const
{
if(x==b.x && y==b.y) return true;
return false;
}
void output()
{
write(x),putchar(' '),write(y),putchar(' '),write(dir),putchar('\n');
}
};
int dis[1010][1010][4],num[1010][1010][4];
char ma[1010][1010],ss[22];
int idx[200],flag,ans;
bool in(int x,int y)
{
return x>=0 && x<n && y>=0 && y<m && ma[x][y]=='.';
}
void bfs(Node s,Node t)
{
queue<Node > q;
int f[3];
while(!q.empty()) q.pop();
dis[s.x][s.y][s.dir]=0;
num[s.x][s.y][s.dir]=1;
q.push(s);
while(!q.empty())
{
Node now=q.front();q.pop();
int x=now.x,y=now.y;
f[0]=now.dir;f[1]=(f[0]+1)%4;f[2]=(f[0]+3)%4;
int xx=x+dir[f[0]][0];
int yy=y+dir[f[0]][1];
if(now==t)
{
flag=1;
for(int i=0;i<4;i++)
if(dis[x][y][i]==now.step)
ans=(ans+num[x][y][i])%mod;
break;
}
if(in(xx,yy))
{
if(dis[xx][yy][f[0]]>now.step+1)
{
dis[xx][yy][f[0]]=now.step+1;
num[xx][yy][f[0]]=num[x][y][f[0]];
q.push(Node(xx,yy,f[0],dis[xx][yy][f[0]]));
}
else if(dis[xx][yy][f[0]]==now.step+1)
num[xx][yy][f[0]]=(num[xx][yy][f[0]]+num[x][y][f[0]])%mod;
}
for(int i=1;i<=2;i++)
{
if(dis[x][y][f[i]]>now.step+1)
{
dis[x][y][f[i]]=now.step+1;
num[x][y][f[i]]=num[x][y][f[0]];
q.push(Node(x,y,f[i],dis[x][y][f[i]]));
}
else if(dis[x][y][f[i]]==now.step+1)
num[x][y][f[i]]=(num[x][y][f[i]]+num[x][y][f[0]])%mod;
}
}
}
int main()
{
// freopen("data.txt","r",stdin);
int cas=1;
idx['N']=0;idx['E']=1;
idx['S']=2;idx['W']=3;
while(read(n)&&read(m)&&read(mod)&&(mod))
{
CLR(dis,INF);CLR(num,0);
Node s,t;flag=ans=0;
for(int i=0;i<n;i++)
scanf("%s",ma[i]);
read(s.x),read(s.y);
read(t.x),read(t.y);
scanf("%s",ss);
s.dir=idx[ss[0]];s.step=0;
bfs(s,t);
if(!flag)
ans=-1;
printf("Case %d: %d %d\n",cas++,mod,ans);
}
return 0;
}