kuangbin专题一 简单搜索

写在前面：

        最近还是想着找专题提升一下自己，尝试来写kuangbin专题，每天更新一点，希望可以更完

 

1、棋盘问题

      用 ban ，visr，visl三个数组记录那些位置可以被访问，然后dfs搜索即可

   

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
typedef pair<int,int> Pi;
typedef unsigned long long ULL;
int gcd(int a,int b){if (b == 0) return a; return gcd(b , a%b);}
int lcm(int a, int b){ return a/gcd(a,b)*b;}
inline int read(){
    int f = 1, x = 0;char ch = getchar();
    while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();}
    while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
const int maxn = 105;
int visr[maxn],visl[maxn];
int ban[maxn][maxn];
int ans,n,k,tot;
char s[maxn][maxn];
void init(){
    memset(visr, 0, sizeof(visr));
    memset(visl, 0, sizeof(visl));
    memset(ban, 0, sizeof(ban));
    ans = 0;
}
void dfs(int x,int y){
    //cout << x << ' ' << y << ' ' << k << endl;
    if (k == 0) {ans++; return;}
    if (x > n || y > n) return;
    if (!visr[x] && !visl[y] && !ban[x][y]){
        visr[x] = visl[y] = 1; k--;
        if (y == n) dfs(x+1, 1);
        else dfs(x, y+1);
        visr[x] = visl[y] = 0; k++;
    }
    if (y == n) dfs(x+1, 1);
    else dfs(x, y+1);
}
int main(){
    //freopen("/Users/chutong/data.txt", "r", stdin);
    while (scanf("%d%d",&n,&k) == 2) {
        if (n == -1 && k == -1) break; init();
        for (int i=1; i<=n; i++){
            scanf("%s",s[i]+1);
            for (int j=1; j<=n; j++) {
                if (s[i][j] == '.') ban[i][j] = 1;
                else tot++;
            }
        }
        dfs(1, 1);
        printf("%d\n",ans);
    }
    return 0;
}

 

2、Dungeon Master

      这是一个三维的搜索问题，在二维平面的基础上，加一个轴，模版不变，判出界，判节点能否访问（bfs比较好，dfs需要剪枝不然太慢）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
int Gcd(int a,int b){if (b == 0) return a; return Gcd(b , a%b);}
int Lcm(int a, int b){ return a/Gcd(a,b)*b;}
inline int read(){
    int f = 1, x = 0;char ch = getchar();
    while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();}
    while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
const int N = 50;
char mp[N][N][N];
int vis[N][N][N];
int dx[N] = {-1,1,0,0,0,0};
int dy[N] = {0,0,1,-1,0,0};
int dz[N] = {0,0,0,0,-1,1};
int L,R,C;
struct Point{
    int x,y,z;
    int step;
};
bool check(int x,int y,int z){
    if (x < 1 || x > R || y < 1 || y > C || z < 1 || z > L) return false;
    return true;
}
int bfs(int sx,int sy,int sz){
    queue<Point> q;
    q.push((Point){sx,sy,sz,0});
    while(!q.empty()){
        Point now = q.front(); q.pop();
        if (mp[now.z][now.x][now.y] == 'E') return now.step;
        for(int i=0; i<6; i++){
            int x = now.x+dx[i],y = now.y+dy[i],z = now.z+dz[i];
            if (!vis[z][x][y] && mp[z][x][y] != '#' && check(x,y,z)){
                vis[z][x][y] = 1;
                q.push((Point){x,y,z,now.step+1});
            }
        }
    }
    return -1;
}
// mp[z][x][y];
int main(){
  //  freopen("/Users/chutong/ACM/data.in","r",stdin);
    while(scanf("%d%d%d",&L,&R,&C) && (L != 0 || R != 0 || C != 0)){
        memset(vis,0,sizeof(vis));
        memset(mp,0,sizeof(mp));
        int sx = 0,sy = 0,sz = 0;
        for(int i=1; i<=L; i++){
            for(int j=1; j<=R; j++){
                scanf("%s",mp[i][j]+1);
                for(int k=1; k<=C; k++){
                    if (mp[i][j][k] == 'S'){
                        sx = j; sy = k; sz = i;
                    }
                }
            }
        }
        int ans = bfs(sx,sy,sz);
        if (ans == -1) printf("Trapped!\n");
        else printf("Escaped in %d minute(s).\n",ans);
    }
    return 0;
}

 

3、Catch That Cow

      后期再尝试下搜索，这题可以转化为图论，i 和 i+1 两个节点建边权为 1，i 和 i*2 建边，最后跑一遍最短路即可（dij单源）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
typedef long long LL;
typedef pair<int,int> Pi;
typedef unsigned long long ULL;
int Gcd(int a,int b){if (b == 0) return a; return Gcd(b , a%b);}
int Lcm(int a, int b){ return a/Gcd(a,b)*b;}
inline int read(){
    int f = 1, x = 0;char ch = getchar();
    while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();}
    while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
const int maxn = 2e6 + 10;
int head[maxn],tot;
int dis[maxn];
struct _edge{
    int to,nex,w;
}edge[maxn];
void add(int u,int v,int w){
    edge[tot] = (_edge){v,head[u],w};
    head[u] = tot++;
}
void init(){
    memset(head,-1,sizeof(head));
}
void dij(int s,int dis[]){
    for(int i=0; i<maxn; i++) dis[i] = 1e9; dis[s] = 0;
    priority_queue<Pi,vector<Pi>,greater<Pi> > pq; 
    pq.push(make_pair(dis[s],s));
    while(!pq.empty()){
        Pi now = pq.top(); pq.pop();
        int u = now.second;
        for(int i=head[u];~i; i=edge[i].nex){
            int v = edge[i].to;
            if (dis[v] > dis[u] + edge[i].w){
                dis[v] = dis[u] + edge[i].w;
                pq.push(make_pair(dis[v],v));
            }
        }
    }
}
int main(){
    int n = read(),k = read();   
    init();
    for(int i=0; i<=1e5; i++){
        add(i,i+1,1); 
        add(i,i-1,1); 
    }
    for(int j=0; j<=1e5; j++){
        add(j,j*2,1);
    }
    dij(n,dis);
    cout << dis[k] << endl;
    return 0;
}