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自然万物都趋向从有序变得无序

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Biridian Forest CodeForces - 329B (逆向bfs)

题目链接:点我

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:

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#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;
}