广搜搜索 WA

地址：http://vjudge.net/contest/view.action?cid=49557#overview

A题：

#include <iostream>#include <queue>using namespace std;#define INF 100005int main(){    int n = 0, k = 0;    cin >> n >> k;    queue<int> q;    q.push (n);    int dist[100005] = {INF};    for (int i=0; i<INF; i++)        dist[i] = INF;    dist[n] = 0;    while (!q.empty()){        int temp = q.front();        //cout << temp <<endl;        //if (temp == k) break;        q.pop();        if (temp-1 > -1 && dist[temp - 1] > dist[temp]+1){<span style="white-space:pre"></span>//①            dist[temp - 1] = dist [temp] + 1;            q.push (temp - 1);            if (temp-1 == k ) break;<span style="white-space:pre"></span>//②        }        if (temp+1 < INF && dist[temp + 1] > dist[temp]+1){            dist[temp + 1] = dist [temp] + 1;            q.push (temp + 1);            if (temp+1 == k ) break;        }        if (2*temp < INF && dist[2*temp] > dist[temp]+1){            dist[2*temp] = dist [temp] + 1;            q.push (2*temp);            if (temp-1 == k ) break;        }    }    cout << dist[k] << endl;}

①：没加判断条件，所有子节点全部插入队列，导致结果错误以及内存超限；

②：当搜索到K点时就可以跳出来，刚开始没加该条件，导致超时。

B题：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int ans = 1;bool a[20][20];int m = 0, n = 0;void dfs (int i, int j);int main(){    while (cin >> m >> n && m && n){        char s[20]={'\0'};        ans = 1;        memset (a, false, sizeof (a));        int loci = 0, locj = 0;        for (int i=0; i<n; i++){            scanf ("%s", s);            for (int j=0; j<m; j++){                if (s[j] == '.') a[i][j] = true;                else if (s[j] == '#') a[i][j] = false;                else a[i][j] = false, loci = i, locj = j;            }        }        //out << loci <<" "<<locj <<endl;        dfs (loci, locj);        cout << ans << endl;    }    return 0;}void dfs (int i, int j){    if (i-1 > -1 && a[i-1][j]){        ans ++;        a[i-1][j] = false;        dfs (i-1 ,j);        //cout << i-1 <<" "<<j <<endl;    }    if (i+1 < n && a[i+1][j]){        ans ++;        a[i+1][j] = false;        dfs (i+1,j);        //cout << i+1 <<" "<<j <<endl;    }    if (j-1 > -1 && a[i][j-1]){<span style="white-space:pre"></span>//①        ans ++;        a[i][j-1] = false;        dfs (i, j-1);        //cout << i <<" "<<j-1 <<endl;    }    if (j+1 < m && a[i][j+1]){<span style="white-space:pre"></span>//②        ans ++;        a[i][j+1] = false ;        dfs (i, j+1);        //cout << i <<" "<<j+1 <<endl;    }}

①②：SB错误，作死典范，，直接copy下来，忘了改值了，竟然还错了两次。。。

C题：

#include <iostream>#include <cstdio>#include <queue>#define INF 100000using namespace std;inline bool inbond (int x) {return (0 <= x && x < 8);}int dist[100] = {0};queue<int> q;void bfs (int xyb);int main(){    char A[4], B[4];    while (scanf ("%s %s",A, B) !=EOF){        //memset (dist, INF, 100);                                                             ① memset()的使用        for (int i=0; i<100; i++)            dist[i] = INF;        int xya = (A[0]-'a')*10 + (A[1]-'1'), xyb = (B[0]-'a')*10 + (B[1]-'1');        //cout <<xya<<xyb<<endl;        while (!q.empty()) q.pop();        q.push (xya);        dist[xya] = 0;        //cout <<xyb<<endl<< dist[xyb] << endl;        bfs (xyb);        cout << "To get from " << A << " to " << B << " takes " << dist[xyb] << " knight moves." <<endl;    }    return 0;}void bfs (int xyb){    int a[2] = {1, -1}, b[2] = {2, -2};    while (!q.empty()){            int temp = q.front ();            q.pop();            int x = temp/10, y = temp % 10;            for (int i=0; i<2; i++){                for (int j=0; j<2; j++){                    if (inbond(x+a[i]) && inbond (y+b[j]) && dist[(x+a[i])*10 + (y+b[j])] > dist[temp] + 1){                        dist[(x+a[i])*10 + (y+b[j])] = dist[temp] + 1;                        //cout << dist[temp] << endl;                        if ((x+a[i])*10 + (y+b[j]) == xyb) return ;                        q.push((x+a[i])*10 + (y+b[j]));                    }                }            }            for (int i=0; i<2; i++){                for (int j=0; j<2; j++){                    if (inbond(x+b[i]) && inbond (y+a[j]) && dist[(x+b[i])*10 + (y+a[j])] > dist[temp] + 1){                        dist[(x+b[i])*10 + (y+a[j])] = dist[temp] + 1;                        //cout << dist[temp] << endl;                        if ((x+b[i])*10 + (y+a[j]) == xyb) return ;                        q.push((x+b[i])*10 + (y+a[j]));                    }                }            }    }}

这道题虽然过了，但是花了好长时间才写好，对广搜还不熟悉；

①：这里memset使用错误，导致一直出错。

     memset()是8位8位赋值，所以，给int 赋值1 时，实际赋的值是0x 01010101, 所以memset 一般只能赋0 或 -1， 可以用memset(dist, 127, sizeof(dist))赋极大值（32639），一般情况就够用了。

D题

#include <iostream>#include <queue>#include <cstring>using namespace std;typedef struct tagPOINT{    int x,y;    int step;}point;int sq;int visited[400][400];void bfs (point now, point tar);inline bool inbond (int a){    return (0 <= a && a < sq);}int main (){    int T = 0;    cin >> T;    while (T--){        memset (visited, 0, sizeof(visited));        point A,B;        cin >> sq;        cin >> A.x >> A.y >> B.x >> B.y;        A.step = 0;        if (A.x == B.x && A.y == B.y){            cout << 0 <<endl;            continue;        }        B.step = sq * sq;        visited[A.x][A.y] = 1;        bfs (A, B);    }    return 0;}void bfs (point now, point tar){    int dir[8][2]={{-2,1},{-2,-1},{-1,2},{-1,-2},{1,2},{1,-2},{2,1},{2,-1}};    queue <point> q;    q.push(now);    while (!q.empty()){        point temp = q.front();                                         //①        q.pop();        for (int i=0; i<8; i++){            point p;            p.x = temp.x + dir[i][0];            p.y = temp.y + dir[i][1];            //cout << p.x << " " << p.y << endl;            if (inbond(p.x) && inbond(p.y) && (visited[p.x][p.y] == 0)){                p.step = temp.step + 1;                if (p.x == tar.x && p.y == tar.y){                    cout << p.step << endl;                    return ;                }                q.push (p);                visited[p.x][p.y] = 1;                                  //②            }        }    }}

①②：延续手残风格，①处定义temp但没给值，②处忘记标记到过的地方；

E题：

  e题直接过 ，就不复制代码了，但是这道题花了好长时间0.0

F题：

#include <iostream>#include <cstring>#include <queue>using namespace std;int grid[25][25], vis[25][25];                              //①const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};const int dy[8] = {-1, 1, 0, 0, 1 ,-1, 1, -1};int ans;int m, n;void add (int x, int y);void bfs (int x, int y);inline bool inbond(int x, int y){    return (0 <= x && x < m && 0 <= y && y < n);}int main(){    int x, y;    while (cin >> m >> n >> x >> y && m && n && x && y){        char stemp[25] = {'\0'};        for (int i=0; i<m; i++){            cin >> stemp;            for (int j=0; j<n; j++){                if (stemp[j] == '.') grid[i][j] = 0;                else grid[i][j] = 1;            }        }        memset(vis, 0, sizeof(vis));        ans = 0;        if (!grid[x-1][y-1]){                                   //②            cout << ans << endl;            continue;        }        bfs(x-1, y-1);        cout << ans << endl;    }}void bfs (int x, int y){    queue<int> q;    q.push(x*n+y);    vis[x][y] = 1;    add(x,y);    while (!q.empty()){        int temp = q.front();        q.pop();        x = temp /n, y = temp % n;        for (int i=0; i<8; i++){            int nx = x + dx[i];            int ny = y + dy[i];            if (inbond(nx, ny) && grid[nx][ny] && !vis[nx][ny]){                add(nx, ny);                q.push(nx*n+ny);                vis[nx][ny] = 1;            }        }    }}void add (int x, int y){    int nx, ny;    ans += 4;    for (int i=0; i<4; i++){        nx = x + dx[i];        ny = y + dy[i];        if (inbond(nx, ny) && grid[nx][ny])            ans--;    }}

① ：该处数组刚开始 只开了20，刚好是界限，结果导致WA了好多次也找不到原因；

②：修改的时候加上了这个判断条件，不知道有没有用。