文章目录
- 双向广搜
- 例题
- 题意
- 分析
- 代码
- 小结
双向广搜
什么是双向广搜？ 如果把bfs想象成在平静的池塘丢一颗石头，激起的波浪一层层扩散到整个空间直到到达目标，就得到起点到终点的最优路径。那么双向广搜就是在起点和终点同时丢石头，两个波浪将在中间某个位置相遇，即得到最优路径。
例题
传送门: HDU-1401
Solitaire is a game played on a chessboard 8x8. The rows and columns of the chessboard are numbered from 1 to 8, from the top to the bottom and from left to right respectively. There are four identical pieces on the board. In one move it is allowed to: move a piece to an empty neighboring field (up, down, left or right), jump over one neighboring piece to an empty field (up, down, left or right).
>There are 4 moves allowed for each piece in the configuration shown above. As an example let’s consider a piece placed in the row 4, column 4. It can be moved one row up, two rows down, one column left or two columns right. Write a program that: reads two chessboard configurations from the standard input, verifies whether the second one is reachable from the first one in at most 8 moves, writes the result to the standard output.
input:
Each of two input lines contains 8 integers a1, a2, …, a8 separated by single spaces and describes one configuration of pieces on the chessboard. Integers a2j-1 and a2j (1 <= j <= 4) describe the position of one piece - the row number and the column number respectively. Process to the end of file.
output:
The output should contain one word for each test case - YES if a configuration described in the second input line is reachable from the configuration described in the first input line in at most 8 moves, or one word NO otherwise.
Sample Input:
代码语言：javascript复制4 4 4 5 5 4 6 5
2 4 3 3 3 6 4 6Sample Output:
代码语言：javascript复制YES题意
在8*8棋盘有四颗棋子，输入开始状态和目标状态的棋子坐标，每次可以移动一个棋子也可以跳过另一个棋子，问能否在8步内走到目标状态。
分析
用双向广搜求解，8个坐标值位压缩成为10进制作为hash值（或者8维数组？）并用unordered_set判重，当hash值出现在另一个分支即相遇。注意剪枝：每个棋子最多只能走4步，因为如果多于4步，那么他们步数之和就会大于8。
代码
代码语言：javascript复制#include <bits/stdc .h>
#include <unordered_set>
using namespace std;
typedef long long ll;
int stepx[4] = { -1,1,0,0 };
int stepy[4] = { 0,0,-1,1 };
struct node {
int p[4], vis;
bool check(int v) { //查重
for (int i : p)
if (i == v)
return 1;
return 0;
}
int hash() {
int a[4];
memcpy(a, p, sizeof(a));
sort(a, a 4);
int res = 0;
for (int i : a)
res = res * 100 i;
return res;
}
};
unordered_set<int> st[2]; //存hashc
queue<node> q[2];
void clear() {
for (int i = 0; i < 2; i ) {
queue<node>c;
swap(q[i], c);
st[i].clear();
}
}
bool ans(int i, int hash) {//队列下标和待查hash
if (st[i].find(hash) != st[i].end()) //相遇
return true;
return false;
}
bool dbfs() {
while (!q[0].empty() || !q[1].empty()) {
for (int i = 0; i < 2; i) { //2个队列
if (!q[i].empty()) {
node cur = q[i].front(); q[i].pop();
cur.vis ;
for (int j = 0; j < 4; j) { //4个点
int x = cur.p[j] / 10, y = cur.p[j] % 10;
for (int k = 0; k < 4; k) { //4个方向
int nx = x stepx[k], ny = y stepy[k];
if (nx >= 1 && nx <= 8 && ny >= 1 && ny <= 8 && cur.check(nx * 10 ny))
nx = stepx[k], ny = stepy[k];
if (nx >= 1 && nx <= 8 && ny >= 1 && ny <= 8 && cur.vis <= 4 && //小于四步
!cur.check(nx * 10 ny)) {
cur.p[j] = nx * 10 ny;
if (st[i].find(cur.hash()) == st[i].end()) {
if (ans(!i, cur.hash()))return true;
q[i].push(cur), st[i].insert(cur.hash()); //入队
}
}
}
cur.p[j] = x * 10 y; //回溯
}
}
}
}
return false;
}
int main() {
while (true) {
clear();
node t;
int a;
t.vis = 0;
for (int k = 0; k < 2; k) {
for (int i = 0; i < 4; i) {
t.p[i] = 0;
for (int j = 0; j < 2; j) {
if (scanf("%d", &a) == EOF)exit(0);
t.p[i] = t.p[i] * 10 a;
}
}
q[k].push(t);
st[k].insert(t.hash());
}
if (ans(0, q[1].front().hash()) || dbfs())
printf("YESn");
else printf("NOn");
}
return 0;
}小结
1.如果爆栈：考虑在入队前做答案检查，出队才检查会多一轮消耗。 2.如何保存中间状态？位压缩.位运算.unordered_set…数组？ 3.String判重TLE让你怀疑人生，string复制会超时,还是老老实实哈希或者康托展开什么的。 4.神奇剪枝起死回生，一定要注意考虑剪枝，因题制宜。
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