Biridian Forest CodeForces

题目链接:点我

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题意:

实际上就是问有多少人可以在开始位置S之前到达终点.

思路:

逆向bfs,记录从终点开始到每个点的距离,然后找出所有的距离比起点离终点小的点,把他们的人数加起来即可.

代码:

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>using namespace std;const int maxn = 1000+5;char w[maxn][maxn];bool vis[maxn][maxn];int d[maxn][maxn];struct ss{    int x, y;}q[maxn * 1000];int sx, sy, tx, ty ,ans, n, m;int dx[]={1,-1,0,0};int dy[]={0,0,1,-1};bool judge(int x,int y){    if(x > 0 && y > 0 && x <= n && y <= m &&!vis[x][y])        return true;    return false;}int bfs(){    int head=0,tail =1;    vis[tx][ty] = true;    q[0].x = tx;    q[0].y = ty;    d[tx][ty] = 0;    while(head<tail){        ss p = q[head++];        int step = d[p.x][p.y] + 1;        for(int  i = 0; i < 4; ++i){            int x = p.x + dx[i];            int y = p.y + dy[i];            if(judge(x, y)){                q[tail].x = x;                q[tail++].y = y;                vis[x][y] = true;                d[x][y] = step;            }        }    }    return 0;}int main(){    memset(vis,false,sizeof(vis));    memset(d,127,sizeof(d));    scanf("%d %d", &n, &m);    for(int i = 1; i <= n; ++i)    for(int j = 1; j <= m; ++j){        scanf(" %c", &w[i][j]);        if(w[i][j] == 'E')            tx = i, ty = j;        if(w[i][j] == 'S')            sx = i, sy = j;        if(w[i][j] == 'T')            vis[i][j] = true;    }ans =0;    bfs();    for(int i = 1; i <= n; ++i)        for(int j = 1; j <= m; ++j)        if(d[i][j] <= d[sx][sy] && w[i][j] >='0' && w[i][j] <= '9')            ans += w[i][j]- '0';    printf("%d\n",ans);    return 0;}