Codeforces 329B Biridian Forest(BFS)

题目链接：Codeforces 329B Biridian Forest

BFS。

不难想到，若小队a距出口的最小距离小于等于主人公距出口的最小距离，小队a一定可一追到主人公，所以只要以出口为起点，做一个bfs记录下出口到各点的距离，之后统计人数即可。需要注意的一点是有些区域可能会被障碍物个离开，初始化的时候要将所有区域到出口的距离设为无穷大，不然在统计时可能会把隔了区的小队人数也统计进去。

#include <iostream>
#include <queue>
#include <cstring>

using namespace std;

const int MAX_N = 1000 + 100;
const int fx[5]={0,0,1,0,-1};
const int fy[5]={0,1,0,-1,0};
const int inf = (1 << 30);
int dis[MAX_N][MAX_N];
bool vis[MAX_N][MAX_N];
char _map[MAX_N][MAX_N];
struct Node
{
    int x,y,num;
};
struct Point
{
    int x,y,dis;
};
Node node[MAX_N * MAX_N];
Node st,en;//start,end
int r,c;

void BFS()
{
    queue<Point> point;
    Point p_en;
    p_en.x = en.x;
    p_en.y = en.y;
    p_en.dis = 0;
    point.push(p_en);
    vis[p_en.x][p_en.y] = true;
    while(!point.empty())
    {
        Point q = point.front();
        point.pop();
        Point _new;
        for(int i = 1;i <= 4;i++)
        {
            _new.x = q.x + fx[i];
            _new.y = q.y + fy[i];
            if(_new.x <= r && _new.x > 0 && _new.y <= c && _new.y > 0 && _map[_new.x][_new.y] != 'T' && !vis[_new.x][_new.y])
            {
                _new.dis = q.dis + 1;
                vis[_new.x][_new.y] = true;
                point.push(_new);
                dis[_new.x][_new.y] = _new.dis;
            }
        }
    }
}
int main()
{
    memset(vis,0,sizeof(vis));
    int cnt = 0;
    cin >> r >> c;
    for(int i = 1;i <= r;i++)
    {
        for(int j = 1;j <= c;j++)
        {
            cin >> _map[i][j];
            dis[i][j] = inf;
            if(_map[i][j] == 'S')
            {
                st.x = i;
                st.y = j;
            }
            else if(_map[i][j] == 'E')
            {
                en.x = i;
                en.y = j;
            }
            else if(_map[i][j] >= '0' && _map[i][j] <= '9')
            {
                cnt++;
                node[cnt].x = i;
                node[cnt].y = j;
                node[cnt].num = _map[i][j] - '0';
            }
        }
    }
    BFS();
    int ans = 0;
    for(int i = 1;i <= cnt;i++)
    {
        if(dis[node[i].x][node[i].y] <= dis[st.x][st.y])
            ans += node[i].num;
    }
    cout << ans << endl;
    return 0;
}