Hrbust 1143 (bfs)

来源：互联网发布：淘宝网购买流程编辑：程序博客网时间：2024/06/03 19:12

题意：

给出n*m的矩形地图，从一个原点出发找出小于等于原点的个数。

思路：

其实可以用深搜去写这道题，不过可以练习一下bfs的写题思路，注意点就是原点
的判断（check）和细节，比如我经常把s，t写错。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int MAXN = 1005;struct P{    int x,y;};int n,m,s,t;int map[MAXN][MAXN];bool vis[MAXN][MAXN];int dir[4][2] = {0,-1,0,1,1,0,-1,0};int check(P a){    if(a.x >= 0 && a.x < n && a.y >= 0 && a.y < m &&        vis[a.x][a.y] != 1 && map[a.x][a.y] <= map[s-1][t-1])        return true;    else        return false;}int solve(){    memset(vis,0,sizeof(vis));    queue<P>que;    P now,temp;    now.x = s-1;    now.y = t-1;    que.push(now);    vis[s-1][t-1] = true;    int ans = 1;    while(!que.empty()) {        temp = que.front();        que.pop();        //vis[tmep.x][temp.y] = true;        for(int i = 0;i < 4; i++) {            now.x = temp.x + dir[i][0];            now.y = temp.y + dir[i][1];            if(check(now)) {                que.push(now);                ans++;                vis[now.x][now.y] = true;            }        }    }    return ans;}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d%d%d%d",&n,&m,&s,&t) != EOF) {        for(int i = 0;i < n; i++) {            for(int j = 0;j < m; j++) {                scanf("%d",&map[i][j]);            }        }        int ans = solve();        printf("%d\n",ans);    }    return 0;}

bfs写法，注意：dfs的入口和出口

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN = 1005;int ans;int n,m,s,t;int map[MAXN][MAXN];int vis[MAXN][MAXN];int dir[4][2] ={0,1,0,-1,1,0,-1,0};void dfs(int a,int b){    ans++,vis[a][b] = true;    for(int i = 0;i < 4; i++) {        int x = a + dir[i][0];        int y = b + dir[i][1];        if(x >= 1 && x <= n && y >= 1 && y <= m &&            vis[x][y] == false && map[x][y] <= map[s][t])                dfs(x,y);    }}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d%d%d%d",&n,&m,&s,&t) != EOF) {        for(int i = 1;i <= n; i++) {            for(int j = 1;j <= m;j ++)                scanf("%d",&map[i][j]);        }        ans = 0;        memset(vis,0,sizeof(vis));        dfs(s,t);        printf("%d\n",ans);    }    return 0;}

0 0