标签：bfs
Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.13
解析：
广搜必须都是1个时间才能搜，才能保证这个BFS树是等距离向外伸展的，而这个不是等距离的，所以需要一些处理。
1、我的方法是，找到天使后，把时间比下大小，最后输出最小的。需要优化，只这么做的话，会TLE的，如果走过一个格子，这个格子存走过时候的时间，下次再走到这个格子，如果时间比格子里的短，就入队，否则，就不用入队了。20MS。
2. 优先队列+BFS 因为等距离的BFS的话，队列里的time值是从小往大排的，那直接用优先队列就可以了丫 0MS
/****ZOJ 1649****/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <set>
#include <queue>
#include <algorithm>
#define MAXN 210
#define INF 1000000 //不要太大，太大结果错误
#define RST(N)memset(N, 0, sizeof(N))
using namespace std;
struct
{
int x, y;
int step;
int time;
}q[1000000];
char Map[MAXN][MAXN]; //地图
int Min[MAXN][MAXN]; //走到每个位置所需的最小时间
int front, rear, res;
int n, m, Sx, Sy, Ex, Ey;
const int dx[] = {-1, 0, 1, 0}; //方向数组
const int dy[] = {0, 1, 0, -1};
bool check(int x, int y) //走下一步的条件
{
return x>=0 && y>=0 && x<n && y<m && Map[x][y]!='#';
}
void Init()
{
front = rear = 0;
memset(Min, INF, sizeof(Min)); //初始化Min
Min[Sx][Sy] = 0;
q[rear].x = Sx, q[rear].y = Sy;
q[rear].time = 0, q[rear++].step = 0; //将起始position加入队列
}
int main(int argc, char *argv[])
{
while(~scanf("%d %d", &n, &m)) {
for(int i=0; i<n; i++) {
scanf("%s", Map[i]);
for(int j=0; Map[i][j]; j++) {
if(Map[i][j] == 'a') { Ex=i, Ey=j; } //记录终点position
else if(Map[i][j] == 'r') { Sx=i, Sy=j; } //记录起始position
}
}
Init(); //个别数据初始化
while(front < rear) { //当队列不为空
int px = q[front].x;
int py = q[front].y; //current position
for(int i=0; i<4; i++) {
int xx = px + dx[i];
int yy = py + dy[i];
//printf("xx = %d yy = %d\n", xx, yy);
if(check(xx, yy)) {
//printf("Matched xx = %d, yy = %d\n", xx, yy);
//v[xx][yy] = 1;
int qt = q[front].time + 1;
if(Map[xx][yy] == 'x') qt++; //如果是警卫，杀死，时间+1
if(qt < Min[xx][yy]) { //如果这种走法比之前走到该位置所花的时间少，则加入队列，否则不用加入队列
Min[xx][yy] = qt;
q[rear].x = xx, q[rear].y = yy;
q[rear].step = q[front].step + 1;
q[rear++].time = qt;
}
//printf("Matched cur_char = %c\n", Map[px][py]);
//printf("Matched next_pos_char = %c\n", Map[xx][yy]);
//printf("Matched cur_step = %d, cur_time = %d\n", q[front].step, q[front].time);
//printf("Matched step = %d, time = %d\n", q[rear].step, q[rear].time);
}
}
front++;
}
if(Min[Ex][Ey] < INF) printf("%d\n", Min[Ex][Ey]);
else puts("Poor ANGEL has to stay in the prison all his life.");
}
return 0;
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <limits.h>
#include <set>
#include <queue>
#include <algorithm>
#define MAXN 210
#define RST(N)memset(N, 0, sizeof(N))
using namespace std;
typedef struct Node
{
int x, y;
int time;
}Node;
Node start;
priority_queue <Node> pq;
char Map[MAXN][MAXN];
int v[MAXN][MAXN], n, m, res;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
bool operator < (Node a, Node b)
{
return a.time > b.time;
}
bool check(int x, int y)
{
return x>=0 && y>=0 && x<n && y<m && Map[x][y]!='#' &&!v[x][y];
}
int BFS()
{
Node cur;
int xx, yy, px, py, T;
while(!pq.empty()) {
cur = pq.top(), pq.pop();
px = cur.x, py = cur.y, T = cur.time;
for(int i=0; i<4; i++) {
xx = px + dx[i], yy = py + dy[i];
if(check(xx, yy)) {
v[xx][yy] = 1;
if(Map[xx][yy] == 'a') return T+1; //基友见面，返回
cur.x = xx, cur.y = yy, cur.time = T+1;
if(Map[xx][yy] == 'x') cur.time++; //打死警卫，时间+1
pq.push(cur); //入队
}
}
}
return -1;
}
int main(int argc, char *argv[])
{
while(~scanf("%d %d", &n, &m)) {
RST(v);
while(!pq.empty()) pq.pop(); //清空队列
for(int i=0; i<n; i++) {
scanf("%s", Map[i]);
for(int j=0; j<m; j++) {
if(Map[i][j] == 'r') { //记录初始position
start.x = i, start.y = j;
start.time = 0;
pq.push(start); //入队
v[i][j] = 1;
}
}
}
res = BFS();
if(res != -1) printf("%d\n", res);
else puts("Poor ANGEL has to stay in the prison all his life.");
}
return 0;
}
ZOJ 1649 && HDU 1242 Rescue （BFS + 优先队列）
标签：bfs
原文地址：http://blog.csdn.net/keshacookie/article/details/44034839
