HDU 4528 (BFS)

来源：互联网发布：知乎中国出口构成编辑：程序博客网时间：2024/05/23 11:41

题目链接：点击这里

题意：给一个地图，要从S出发看到D，E，如果两个点在同一行并且中间没有X或者人就可以看到。

比较坑的是不能走有人的格子。所以直接给两个人一个权值多开一维记录拿到的权值就好了。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>using namespace std;#define maxn 111int n, m, t;char mp[maxn][maxn];bool legal (int x, int y) {    if (x < 0 || y < 0 || x >= n || y >= m)        return 0;    if (mp[x][y] == 'X' || mp[x][y] == 'E' || mp[x][y] == 'D')        return 0;    return 1;}struct node {    int x, y;    int num, step;    node (int _x = 0, int _y = 0, int _n = 0, int _s = 0):x(_x), y(_y), num(_n), step(_s) {}}s, e, d;queue <node> gg;int val[maxn][maxn];void go (node s, int num) {     int x, y;    x = s.x, y = s.y;    for (y--; legal (x, y); y--) {val[x][y] |= num;} y = s.y;    for (y++; legal (x, y); y++) {val[x][y] |= num;} y = s.y;    for (x--; legal (x, y); x--) {val[x][y] |= num;} x = s.x;    for (x++; legal (x, y); x++) {val[x][y] |= num;} x = s.x;}bool vis[maxn][maxn][11];const int M[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};int bfs () {    while (!gg.empty ()) gg.pop ();    memset (val, 0, sizeof val);    go (e, 1);    go (d, 2);    s.num += val[s.x][s.y];    memset (vis, 0, sizeof vis);    vis[s.x][s.y][s.num] = 1;    gg.push (s);    while (!gg.empty ()) {        node now = gg.front (); gg.pop ();        if (now.num == 3) {            return now.step;        }        for (int i = 0; i < 4; i++) {            node next = now;            next.x += M[i][0], next.y += M[i][1];            if (!legal (next.x, next.y))                continue;            next.step++;            next.num |= val[next.x][next.y];            if (vis[next.x][next.y][next.num] || next.step > t)                continue;            gg.push (next);            vis[next.x][next.y][next.num] = 1;        }    }    return -1;}int main () {    int T, kase = 0;    scanf ("%d", &T);    while (T--) {        printf ("Case %d:\n", ++kase);        scanf ("%d%d%d", &n, &m, &t);        for (int i = 0; i < n; i++) {            scanf ("%s", mp[i]);            for (int j = 0; j < m; j++) {                if (mp[i][j] == 'S') s = node (i, j, 0);                else if (mp[i][j] == 'D') d = node (i, j, 0);                else if (mp[i][j] == 'E') e = node (i, j, 0);            }        }        int ans = bfs ();        printf ("%d\n", ans);    }    return 0;}

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