HRBUST1143.泉水

BFS解法：

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
struct Point
{
    int x;
    int y;
} tmp,tmp2;
queue<Point>q;
int vis[1005][1005];
int a[1005][1005];
int d[4][2]= {0,1,0,-1,1,0,-1,0};//四個方向
int i,j;
int main()
{
    int n,m,p1,p2;
    while(scanf("%d%d%d%d",&n,&m,&p1,&p2)!=EOF)
    {
        int ans=1;
        for(i=1; i<=n; i++)
            for(j=1; j<=m; j++)
                scanf("%d",&a[i][j]);
        while(!q.empty())//隊列清空
        {
            q.pop();
        }
        memset(vis,0,sizeof(vis));//數組清空
        tmp.x = p1;
        tmp.y = p2;
        vis[tmp.x-1][tmp.y-1]=1;
        q.push(tmp);
        while(!q.empty())
        {
            tmp = q.front();
            q.pop();
            for(i=0; i<4; i++)//各方向搜索
            {
                tmp2.x = tmp.x + d[i][0];
                tmp2.y = tmp.y + d[i][1];
                if(a[tmp2.x][tmp2.y]<=a[p1][p2] && tmp2.x>=1 && tmp2.x<=n && tmp2.y>=1 && tmp2.y<=m && !vis[tmp2.x-1][tmp2.y-1])//入隊條件
                {
                    ans++;//每淹沒一點，答案加一
                    vis[tmp2.x-1][tmp2.y-1] = 1;//標記淹沒點
                    a[tmp2.x][tmp2.y] = a[p1][p2];//淹沒點與泉眼高度一致
                    q.push(tmp2);
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

DFS解法：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int n,m,p1,p2;
int vis[1005][1005];
int a[1005][1005];
int i,j,ans;
int d[4][2]= {0,1, 0,-1, 1,0, -1,0};
void DFS(int x,int y)
{
    if(vis[x][y])
    {
        return ;
    }
    if(!vis[x][y])
    {
        ans++;
        vis[x][y]=1;
    }
    int i,x1,y1;
    for(i=0; i<4; i++)
    {
        x1=x+d[i][0];
        y1=y+d[i][1];
        if(x1>=1 && x1<=n && y1>=1 && y1<=m && a[x1][y1]<=a[p1][p2])
        {
            DFS(x1,y1);//通過遞歸實現深搜
        }
    }
    return ;
}
int main()
{
    while(scanf("%d%d%d%d",&n,&m,&p1,&p2) != EOF)
    {
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        memset(vis,0,sizeof(vis));
        ans=0;
        DFS(p1,p2);
        printf("%d\n",ans);
    }
}