大一集训 搜索技术科普

A - Red and Black HDU - 1312  dfs

题目大意：给你起点，障碍物不可越过，最多能去到几个点。

#include <iostream>
#include <map>
#include<cstdio>
using namespace std;
int sx, sy,w,h,ans=0;
char mp[21][21];
int vis[21][21];
void dfs(int x, int y) {
    if (x < 0 || y < 0 || x >= h || y >= w || mp[x][y] == '#'|| vis[x][y] == 1)return;
    ans++; vis[x][y] = 1;
    dfs(x-1, y);dfs(x+1, y);dfs(x, y-1);dfs(x, y+1);
}
int main()
{
    while (scanf("%d %d", &w, &h), w + h) {
        for (int i = 0;i < h;i++) {
            scanf("%s", mp[i]);
            for (int j = 0;j < w;j++) {
                if (mp[i][j] == '@') { sx = i, sy = j; }
            }
        }
        dfs(sx, sy);
        printf("%d\n", ans);
        ans = 0;
        memset(vis, 0, sizeof vis);
    }
}

 C - N皇后问题 HDU - 2553 

#include <iostream>
#include <map>
#include <string>
#include<algorithm>
#include<vector>
#include<cmath>
#include<queue>
#define max 20
using namespace std;
int n, ans[max], mp[max][max], res;
bool ok(int x, int y) {
    for (int i = 1;i <= x - 1;i++) {
        for (int j = 1;j <= n;j++) {
            if (!(x == i && y == j)) {//除去需要判断的点
                if (mp[i][j] == 1) {//该点放有皇后
                    if (j == y || i + j == x + y || i - j == x - y)return false;//判断列，正负对角线是否满足
                }
            }
        }
    }
    return true;
}
void dfs(int queen) {
    if (queen == n + 1) {
        res++;
        return;
    }
    for (int i = 1;i <= n;i++) {
        if (ok(queen, i)) {
            mp[queen][i] = 1;
            dfs(queen + 1);
            mp[queen][i] = 0;
        }
    }
}
int main() {
    memset(ans, -1, sizeof(ans));
    while (scanf("%d", &n), n) {
        if (ans[n] != -1) {//没有ans会超时，相当于预处理
            printf("%d\n", ans[n]);
            continue;
        }
        memset(mp, 0, sizeof(mp));
        res = 0;
        dfs(1);
        ans[n] = res;
        printf("%d\n", ans[n]);
    }
    return 0;
}

G - Catch That Cow POJ - 3278   bfs

 题目大意：农民抓牛，农民可以选择前进或后退一步，或者达到当前距离的两倍地方。

#include <iostream>
#include <map>
#include <string>
#include<algorithm>
#include<vector>
#include<cmath>
#include<queue>
#define max 100010
int vis[max] = { 0 };
struct Node {
    int x;
    int step;
}temp, nxt;
using namespace std;
int n, k;
int a[max];
int bfs(int a) {
    temp.x = a;
    temp.step = 0;
    queue<Node>q;
    q.push(temp);
    while (!q.empty()) {
        temp = q.front();
        q.pop();
        if (temp.x == k)return temp.step;
        for (int i = 0;i < 3;i++) {
            if (i == 0) { nxt.x = temp.x * 2; }
            else if (i == 1) { nxt.x = temp.x + 1; }
            else if (i == 2) { nxt.x = temp.x - 1; }
            if (nxt.x < 0 || nxt.x>max || vis[nxt.x] == 1)continue;//如果把vis[nxt.x]放前面，nxt.x可能会超过vis数组开的范围，例如（100000*2），或者直接改变vis数组的大小
            nxt.step = temp.step + 1;
            vis[nxt.x] = 1;
            q.push(nxt);
        }
    }
}
int main() {
    while (scanf("%d %d", &n, &k) != EOF) {
        memset(vis, 0, sizeof(vis));
        printf("%d\n", bfs(n));
    }
}

H - Dungeon Master POJ - 2251

三维立体图，从S到E，不能经过‘#’

#include <iostream>
#include <map>
#include <string>
#include<algorithm>
#include<vector>
#include<cmath>
#include<queue>
#define max 10002
using namespace std;
int vis[35][35][35] = { 0 };
int sx, sy, sz;
int ex, ey, ez;
char mp[35][35][35];
int dx[6] = { 1,0,-1,0,0,0 }, dy[6] = { 0,1,0,-1 }, dz[6] = {0,0,0,0,1,-1};
int l, r, c;
typedef struct Node {
    int x, y, z;
    int step;
}node;
int bfs(int a, int b, int cc) {
    node temp, next;
    temp.x = a;
    temp.y = b;
    temp.z = cc;
    temp.step = 0;
    queue<node>q;
    q.push(temp);
    while (!q.empty()) {
        temp = q.front();
        q.pop();
        if (temp.x == ex && temp.y == ey && temp.z == ez)return temp.step;
        for (int i = 0;i < 6;i++) {
            if ((vis[temp.x + dx[i]][temp.y + dy[i]][temp.z + dz[i]] == 1))continue;
            if((mp[temp.x + dx[i]][temp.y + dy[i]][temp.z + dz[i]] == '#'))continue;
            if ((temp.x + dx[i]) < 0)continue;
            if((temp.x + dx[i])>=r)continue;
            if ((temp.y + dy[i])<0)continue;
            if ((temp.y + dy[i])>=c)continue;
            if ((temp.z + dz[i])<0)continue;
            if ((temp.z + dz[i])>=l)continue;
            next.x=temp.x + dx[i];
            next.y=temp.y + dy[i];
            next.z=temp.z + dz[i];
            next.step = temp.step + 1;
            vis[next.x][next.y][next.z] = 1;
            q.push(next);
        }
    }
    return -1;
}
int main() {
    while (scanf("%d %d %d", &l, &r, &c) ,l+r+c) {
        memset(vis, 0, sizeof(vis));
        memset(mp, '0', sizeof(mp));
        getchar();
        for (int i = 0;i < l;i++) {
            for (int j = 0;j < r;j++) {
                for (int k = 0;k < c;k++) {
                    scanf("%c", &mp[j][k][i]);
                    if (mp[j][k][i] == 'S') { sx = j;sy = k;sz = i; }
                    if (mp[j][k][i] == 'E') { ex = j;ey = k;ez = i; }
                }
                getchar();
            }
                if(i!=l-1)getchar();
        }
        int res = bfs(sx, sy, sz);
        if (res== -1)printf("Trapped!\n");
        else printf("Escaped in %d minute(s).\n", res);
    }
}

I - Prime Path HDU - 1973

题意：给你两个四位数，都是素数，每次改变可以改变其中的任何一个数字，但要求改变后的四位数（没有前导零）依然是素数，问最少改变几次可以使得第一个数改为第二个数

#include <iostream>
#include <map>
#include <string>
#include<algorithm>
#include<vector>
#include<cmath>
#include<queue>
#define max 10002
using namespace std;
int n, m;
int vis[max] = { 0 };
int num[max] = { 0 };
typedef struct Node {
    int x;
    int step;
}node;
void getprime() {
    for (int i = 1001;i <= 9999;i += 2) {
        int flag = 1;
        for (int j = 2;j <= sqrt(i);j++) {
            if (i % j == 0) {flag = 0;break;}
        }
        if (flag == 1)num[i] = 1;
    }
}
int bfs(int a) {
    node temp, next;
    temp.x = a;
    temp.step = 0;
    queue<node>q;
    q.push(temp);
    while (!q.empty()) {
        temp = q.front();
        q.pop();
        if (temp.x == m)return temp.step;
        for (int i = temp.x / 1000;i < 9;i++) {
            if (vis[temp.x + 1000 * (i - temp.x / 1000 + 1)] ==1 )continue;
            if (num[temp.x + 1000 * (i - temp.x / 1000 + 1)] == 1) {next.x = temp.x + 1000 * (i-temp.x/1000+1);next.step = temp.step + 1;vis[next.x] = 1;q.push(next);}
        }
        for (int i = temp.x / 1000;i >1;i--) {
            if (vis[temp.x - 1000 * (temp.x / 1000 - i + 1)] == 1)continue;
            if (num[temp.x - 1000 * (temp.x / 1000 - i + 1)] == 1) { next.x = temp.x - 1000 * (temp.x/1000-i+1);next.step = temp.step + 1;vis[next.x] = 1; q.push(next);}
        }
        for (int i = temp.x%1000/100;i < 9;i++) {
            if (vis[temp.x + 100 * (i - temp.x % 1000 / 100 + 1)] == 1)continue;
            if (num[temp.x + 100 * (i - temp.x % 1000 / 100 + 1)] == 1) { next.x = temp.x + 100 * (i- temp.x % 1000 / 100+1);next.step = temp.step + 1;vis[next.x] = 1;q.push(next); }
        }
        for (int i = temp.x%1000/100;i > 0;i--) {
            if (vis[temp.x - 100 * (temp.x % 1000 / 100 - i + 1)] == 1)continue;
            if (num[temp.x - 100 * (temp.x % 1000 / 100 - i + 1)] == 1) { next.x = temp.x - 100 * (temp.x % 1000 / 100-i+1);next.step = temp.step + 1;vis[next.x] = 1; q.push(next); }
        }
        for (int i = temp.x % 100 / 10;i < 9;i++) {
            if (vis[temp.x + 10 * (i - temp.x % 100 / 10 + 1)] == 1)continue;
            if (num[temp.x + 10 * (i - temp.x % 100 / 10 + 1)] == 1) { next.x = temp.x + 10 * (i- temp.x % 100 / 10+1);next.step = temp.step + 1;vis[next.x] = 1;q.push(next); }
        }
        for (int i = temp.x % 100 / 10;i > 0;i--) {
            if (vis[temp.x - 10 * (temp.x % 100 / 10 - i + 1)] == 1)continue;
            if (num[temp.x - 10 * (temp.x % 100 / 10 - i + 1)] == 1) { next.x = temp.x - 10 * (temp.x % 100 / 10-i+1);next.step = temp.step + 1;vis[next.x] = 1; q.push(next); }
        }
        for (int i = temp.x % 10;i < 9;i++) {
            if (vis[temp.x + 1 * (i - temp.x % 10 + 1)] == 1)continue;
            if (num[temp.x + 1 * (i - temp.x % 10 + 1)] == 1) { next.x = temp.x + 1 * (i- temp.x % 10+1);next.step = temp.step + 1;vis[next.x] = 1;q.push(next); }
        }
        for (int i = temp.x % 10;i > 0;i--) {
            if (vis[temp.x - 1 * (temp.x % 10 - i + 1)] == 1)continue;
            if (num[temp.x - 1 * (temp.x % 10 - i + 1)] == 1) { next.x = temp.x - 1 * (temp.x % 10-i+1);next.step = temp.step + 1;vis[next.x] = 1; q.push(next); }
        }
    }
}
int main() {
    int cn;
    scanf("%d", &cn);
    getprime();
    while (cn--) {
        memset(vis, 0, sizeof(vis));
        scanf("%d %d", &n, &m);
        printf("%d\n", bfs(n));
    }
    return 0;
}