暑期个人赛--第一场--D

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时间限制 1000 ms 内存限制 65536 KB

题目描述

给定一个N∗M的矩阵，求问里面有多少个由'#'组成的矩形,"There are 5 ships."，若是里面有一个不是矩形的联通块，则输出"So Sad"

输入格式

1≤n,m≤1000

有多组数据，EOF结束。

输出格式

每行对应一个answer

输入样例

6 8.....#.###.....###.....#.......##......##..#...#6 8.....#.###.....####...##.......###.....##..#...#

输出样例

There are 5 ships.So Sad

赛中提交：T

赛后ac：Y

题目大意：在矩阵里面找出独立的由#组成的区域个数

思路：

裸的广搜

赛中为何不过：

真心不懂- -赛后直接复制赛中提交的再提交一次一样的就过了

代码如下

#include "iostream"#include "string.h"#include "stdio.h"#include "stdlib.h"#include "math.h"#include "string"#include "vector"#include "list"#include "map"#include "queue"#include "stack"#include "bitset"#include "algorithm"#include "numeric"#include "functional" using namespace std;typedef pair<int ,int > P; int n,m;int u,d,l,r,counta,countaa;char c[1005][1005];int vis[1005][1005];int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1}; void bfs(int bx,int by){    P p,tempt;    queue<P> que;    tempt.first=bx,tempt.second=by;    que.push(tempt);    vis[bx][by]=1;    while(que.size()){        int x,y;        counta+=1;        tempt=que.front();        que.pop();        x=tempt.first,y=tempt.second;        //printf("\n\nque.size()=%d  取%d %d\n",que.size(),x,y);        l=min(y,l);        r=max(y,r);        u=max(x,u);        d=min(x,d);        //printf("上%d下%d左%d右%d\n",u,d,l,r);         for(int i=0;i<4;i+=1){            int nx=x+dx[i],ny=y+dy[i];            if(nx>=0&&ny>=0&&nx<n&&ny<m&&c[nx][ny]!='.'&&vis[nx][ny]==0){                tempt.first=nx,tempt.second=ny;                que.push(tempt);                vis[nx][ny]=1;            }        }    }} int main(){    while(scanf("%d %d",&n,&m)!=EOF){        getchar();        int F=0;        countaa=0;        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i+=1){            for(int j=0;j<m;j+=1){                scanf("%c",&c[i][j]);            }            getchar();        }         for(int i=0;i<n;i+=1){            bool flag=0;            for(int j=0;j<m;j+=1){                //    printf("(%d,%d)",i+1,j+1);                if(!vis[i][j]&&c[i][j]=='#'){                    u=d=i;                    r=l=j;                    counta=0;                    bfs(i,j);                    if(counta!=(u-d+1)*(r-l+1)){                        F=1;                        //printf("hhhhh(u%d-d%d+1)*(r%d-l%d+1)=%d  counta=%d\n",u,d,r,l,(u-d+1)*(r-l+1),counta);                        break;                    }                    else{                        countaa+=1;                    }                }                //printf("dfasfd%d %d\n",i+1,j+1);            }            if(F){                //    printf("\n\ncome and eat shit\n\n");                break;            }        }        if(F){            printf("So Sad\n");        }        else{            printf("There are %d ships.\n",countaa);        }    }    return 0;}

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