201312-5

//分别从初始位置和终点出法，找到起点可以到达而终点无法到达的点 
#include <iostream>
#include <algorithm>
#include <string.h>
#include <cmath>
#include <stdio.h>
#include <queue>
#define maxn 60
char mmap[maxn][maxn];
int vis[maxn][maxn];//标记从起点出发
int vis1[maxn][maxn];//标记从终点出发 
int dy[4] = {0,0,-1,1};//转向数组 
int dx[4] = {1,-1,0,0};
int n,m;
int num[maxn][maxn][maxn][maxn];//！！
using namespace std;
struct Node
{
    int x;
    int y;
    Node(){}
    Node (int _x,int _y){
        x = _x;
        y = _y;
    }
 } node[maxn];
 Node st,ed;
 int check(Node a){
     if(a.x>=0 && a.x<n && a.y>=0 && a.y<m && mmap[a.x][a.y]!='#') return 1;
    return 0; 
 } 
 int get(char c,int &u,int &v){//判断当前可走的方向
     if(c == '-')
    {
        u = 2;
        v = 4;
    }
    else if(c == '|')
    {
        u = 0;
        v = 2;
    }
    else if(c == '.')
    {
        u = 0;
        v = 1;
    }
    else
    {
        u = 0;
        v = 4;
    }
}
queue<Node>que;
int  bfs()
{
    while(!que.empty()) {
        que.pop();
    }
    que.push(st);  //从起点走 
    vis[st.x][st.y] = 1;
    Node tmp, now;
    while(!que.empty())
    {
        tmp = que.front();
        que.pop();
        int u, v;
        get(mmap[tmp.x][tmp.y], u, v);
        for(int i = u; i < v; i++)
        {
            now.x = tmp.x + dx[i];
            now.y = tmp.y + dy[i];
            if(check(now))
            {
                num[tmp.x][tmp.y][now.x][now.y] = 1;
                if(!vis[now.x][now.y]) {
                    vis[now.x][now.y] = 1;
                    que.push(now);
                }
            }
        }
    }
    if(vis[ed.x][ed.y] == 0)
    {
        puts("I'm stuck!");// 若未到达 
        return 0;
    }
    que.push(ed);  //从终点走 
    while(!que.empty())
    {
        tmp = que.front();
        que.pop();
        for(int i = 0; i < 4; i++)
        {
            now.x = tmp.x + dx[i];
            now.y = tmp.y + dy[i];
            if(check(now) && vis1[now.x][now.y] == 0 && num[now.x][now.y][tmp.x][tmp.y])
            {
                vis1[now.x][now.y] = 1;
                que.push(now);
            }
        }
    }
    int ans = 0;
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            if(vis[i][j] == 1 && vis1[i][j] == 0)
            {
                ans++;
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}
int main()
{
    scanf("%d %d", &n, &m);
    memset(vis, 0, sizeof(vis));
    memset(vis1, 0, sizeof(vis1));
    memset(num, 0, sizeof(num));
    for(int i = 0; i < n; i++)
    {
        scanf("%s", mmap[i]);
        for(int j = 0; j < m; j++)
        {
            if(mmap[i][j] == 'S')
            {
                st.x = i;
                st.y = j;
            }
            if(mmap[i][j] == 'T')
            {
                ed.x = i;
                ed.y = j;
            }
        }
    }
    bfs();
    return 0;
}

注意：

1.读准条件，能够到达，则从起始点开始寻找。

2.无法到达终点则从结束点反向寻找

3.注意细节，对于正向无法到达的路径，则反向也没有必要去寻找，因此需要int num[maxn][maxn][maxn][maxn];来记录正向寻找过的路径

标签：tmp,int,201312,vis,que,maxn,now
来源： https://www.cnblogs.com/zmachine/p/13768269.html