Nightmare
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7367 Accepted Submission(s): 3530
Problem Description
Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bomb is
set to 6 minutes. To prevent the bomb from exploding by shake, Ignatius had to move slowly, that is to move from one area to the nearest area(that is, if Ignatius stands on (x,y) now, he could only on (x+1,y), (x-1,y), (x,y+1), or (x,y-1) in the next minute)
takes him 1 minute. Some area in the labyrinth contains a Bomb-Reset-Equipment. They could reset the exploding time to 6 minutes.
Given the layout of the labyrinth and Ignatius‘ start position, please tell Ignatius whether he could get out of the labyrinth, if he could, output the minimum time that he has to use to find the exit of the labyrinth, else output -1.
Here are some rules:
1. We can assume the labyrinth is a 2 array.
2. Each minute, Ignatius could only get to one of the nearest area, and he should not walk out of the border, of course he could not walk on a wall, too.
3. If Ignatius get to the exit when the exploding time turns to 0, he can‘t get out of the labyrinth.
4. If Ignatius get to the area which contains Bomb-Rest-Equipment when the exploding time turns to 0, he can‘t use the equipment to reset the bomb.
5. A Bomb-Reset-Equipment can be used as many times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as many times as you wish.
6. The time to reset the exploding time can be ignore, in other words, if Ignatius get to an area which contain Bomb-Rest-Equipment, and the exploding time is larger than 0, the exploding time would be reset to 6.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case starts with two integers N and M(1<=N,Mm=8) which indicate the size of the labyrinth. Then N lines follow, each line contains M integers. The array indicates the layout of the labyrinth.
There are five integers which indicate the different type of area in the labyrinth:
0: The area is a wall, Ignatius should not walk on it.
1: The area contains nothing, Ignatius can walk on it.
2: Ignatius‘ start position, Ignatius starts his escape from this position.
3: The exit of the labyrinth, Ignatius‘ target position.
4: The area contains a Bomb-Reset-Equipment, Ignatius can delay the exploding time by walking to these areas.
Output
For each test case, if Ignatius can get out of the labyrinth, you should output the minimum time he needs, else you should just output -1.
Sample Input
3 3 3 2 1 1 1 1 0 1 1 3 4 8 2 1 1 0 1 1 1 0 1 0 4 1 1 0 4 1 1 0 0 0 0 0 0 1 1 1 1 4 1 1 1 3 5 8 1 2 1 1 1 1 1 4 1 0 0 0 1 0 0 1 1 4 1 0 1 1 0 1 1 0 0 0 0 3 0 1 1 1 4 1 1 1 1 1
Sample Output
4 -1 13
Author
Ignatius.L
题意：给定一张n行m列的地图，其中1代表空地，2代表入口，3代表出口，4代表炸弹时间重置点，炸弹时间初始化时是6秒，每次移动只能是上下左右相邻的格子，每次耗时1秒，0秒时炸弹立刻爆炸，即便到达出口或者重置点也没用，求从入口达到出口所花最少时间，如果出不来输出-1。
题解：用广搜解，由于数据量较少，不需要重复的状态标记，但是为了避免死循环，时间重置点只能用1次，用完后将其标记成空地。
#include <stdio.h>
#include <string.h>
#include <queue>
#define maxn 10
int G[maxn][maxn], n, m;
struct Node {
int x, y, time, leftTime;
} S;
const int mov[][2] = {0, 1, 0, -1, 1, 0, -1, 0};
void getMap() {
scanf("%d%d", &n, &m);
int i, j;
for(i = 1; i <= n; ++i)
for(j = 1; j <= m; ++j) {
scanf("%d", &G[i][j]);
if(G[i][j] == 2) {
S.x = i; S.y = j;
S.time = 0; S.leftTime = 6;
}
}
}
bool check(int x, int y) {
if(x < 1 || x > n || y < 1 || y > m)
return false;
if(G[x][y] == 0) return false;
return true;
}
int BFS() {
int i, j, x, y;
std::queue<Node> Q;
Q.push(S);
Node now, tmp;
while(!Q.empty()) {
now = Q.front(); Q.pop();
for(i = 0; i < 4; ++i) {
x = mov[i][0] + now.x;
y = mov[i][1] + now.y;
if(check(x, y)) {
tmp = now; ++tmp.time;
if(--tmp.leftTime) {
if(G[x][y] == 4) {
tmp.leftTime = 6;
G[x][y] = 1;
} else if(G[x][y] == 3)
return tmp.time;
tmp.x = x; tmp.y = y;
Q.push(tmp);
}
}
}
}
return -1;
}
void solve() {
printf("%d\n", BFS());
}
int main() {
// freopen("stdin.txt", "r", stdin);
int t;
scanf("%d", &t);
while(t--) {
getMap();
solve();
}
return 0;
}
带vis标记的解法：
#include <stdio.h>
#include <string.h>
#include <queue>
#define maxn 10
int G[maxn][maxn], n, m;
struct Node {
int x, y, time, leftTime;
} S;
const int mov[][2] = {0, 1, 0, -1, 1, 0, -1, 0};
bool vis[maxn][maxn][8];
void getMap() {
scanf("%d%d", &n, &m);
int i, j;
for(i = 1; i <= n; ++i)
for(j = 1; j <= m; ++j) {
scanf("%d", &G[i][j]);
if(G[i][j] == 2) {
S.x = i; S.y = j;
S.time = 0; S.leftTime = 6;
}
}
}
bool check(int x, int y) {
if(x < 1 || x > n || y < 1 || y > m)
return false;
if(G[x][y] == 0) return false;
return true;
}
int BFS() {
int i, j, x, y;
std::queue<Node> Q;
Q.push(S);
Node now, tmp;
memset(vis, 0, sizeof(vis));
vis[S.x][S.y][S.leftTime] = 1;
while(!Q.empty()) {
now = Q.front(); Q.pop();
for(i = 0; i < 4; ++i) {
x = mov[i][0] + now.x;
y = mov[i][1] + now.y;
tmp = now;
if(check(x, y) && !vis[x][y][--tmp.leftTime]) {
++tmp.time;
if(tmp.leftTime) {
if(G[x][y] == 4) {
tmp.leftTime = 6;
G[x][y] = 1;
} else if(G[x][y] == 3)
return tmp.time;
tmp.x = x; tmp.y = y;
vis[x][y][tmp.leftTime] = 1;
Q.push(tmp);
}
}
}
}
return -1;
}
void solve() {
printf("%d\n", BFS());
}
int main() {
// freopen("stdin.txt", "r", stdin);
int t;
scanf("%d", &t);
while(t--) {
getMap();
solve();
}
return 0;
}