FZU 2196 Escape (两次bfs)

来源：互联网发布：黑客帝国3矩阵革命编辑：程序博客网时间：2024/05/18 06:48

题意：

小明进入地下迷宫寻找宝藏，找到宝藏后却发生地震，迷宫各处产生岩浆，小明急忙向出口处逃跑。如果丢下宝藏，小明就能迅速离开迷宫，但小明并不想轻易放弃自己的辛苦所得。所以他急忙联系当程序员的朋友你（当然是用手机联系），并告诉你他所面临的情况，希望你能告诉他是否能成功带着宝藏逃脱。

题解：

首先用Spfa预处理出火到达每个位置时间。然后这个人从起点开始bfs，注意一点如果火和人同时到达终点人可以逃脱！！！这个有点坑！手写队列开小了，一直wa。

#include<iostream>#include<math.h>#include<stdio.h>#include<algorithm>#include<string.h>#include<vector>#include<queue>#include<map>#include<set>#define B(x) (1<<(x))using namespace std;typedef long long ll;void cmax(int& a,int b){ if(b>a)a=b; }void cmin(int& a,int b){ if(b<a)a=b; }void cmax(ll& a,ll b){ if(b>a)a=b; }void cmin(ll& a,ll b){ if(b<a)a=b; }void add(int& a,int b,int mod){ a=(a+b)%mod; }void add(ll& a,ll b,ll mod){ a=(a+b)%mod; }const int oo=0x3f3f3f3f;const int MOD=1000000007;const double eps = 1e-9;const int maxn = 1005;int rock[maxn][maxn];char maze[maxn][maxn];int Time[maxn][maxn], vis[maxn][maxn];int n,m;struct Node{    int x, y;    Node(){}    Node(int a, int b){        x = a;        y = b;    }};Node Q[maxn * maxn];int f, r;int d[4][2]={  {1, 0},{0, 1},{-1, 0},{0, -1}};bool ok(int x, int y){    if(x >= 1 && x <= n && y >= 1 && y <= m)        if(maze[x][y] != '!' && maze[x][y] != '#')            return true;    return false;}Node Init(){    scanf("%d %d", &n, &m);    Node s;    memset(rock, 0x3f, sizeof rock);    memset(vis, 0, sizeof vis);    f = r = 0;    for(int i = 1; i <= n; i++){        scanf("%s", maze[i] + 1);        for(int j = 1; j <= m; j++){            if(maze[i][j] == '!'){                Q[r++] = Node(i, j);                vis[i][j] = 1;                rock[i][j] = 0;            }            if(maze[i][j] == 'S')s = Node(i, j);        }    }    Node now;    int x, y;    while(f < r){        now = Q[f++];        vis[now.x][now.y] = 0;        for(int i = 0; i < 4; i++){            x = now.x + d[i][0];            y = now.y + d[i][1];            if(ok(x, y)){                if(rock[now.x][now.y] + 1 < rock[x][y]){                    rock[x][y] = rock[now.x][now.y] + 1;                    if(!vis[x][y]){                        vis[x][y] = 1;                        Q[r++] = Node(x, y);                    }                }            }        }    }    return s;}void bfs(Node s){    memset(vis, 0, sizeof vis);    memset(Time, 0x3f, sizeof Time);    f = r = 0;    Q[r++] = s;    Time[s.x][s.y] = 0;    vis[s.x][s.y] = 1;    Node now;    int x, y;    while(f < r){        now = Q[f++];        for(int i = 0; i <4; i++){            x = now.x + d[i][0];            y = now.y + d[i][1];            if(ok(x, y) && !vis[x][y]){                if(maze[x][y] == '.'){                    if(Time[now.x][now.y] + 1 < rock[x][y])                        if(Time[now.x][now.y] + 1 < Time[x][y]){                            Time[x][y] = Time[now.x][now.y] + 1;                            Q[r++] = Node(x, y);                        }                }else if(maze[x][y] == 'E'){                    if(Time[now.x][now.y] + 1 <= rock[x][y]){                        printf("Yes\n");                        return ;                    }                }            }        }    }    printf("No\n");}int main(){    //freopen("E:\\read.txt","r",stdin);    int T;    scanf("%d", &T);    while(T--){        bfs(Init());    }    return 0;}/*1003 4!S.........E3 4.S...!.....E3 4.S.....#..#E3 4.......S..!E11 3!ES*/

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