POJ 1077 HDU 1043 Eight (IDA*)
题意就不用再说明了吧......如此经典
之前想用双向广搜、a*来写，但总觉得无力，现在用IDA*感觉其他的解法都弱爆了..............想法活跃，时间，空间消耗很小，给它跪了
启发式搜索关键还是找估价函数：此题估价函数可大致定性为每个数字（除去x，只要8个数字）当前位置与它期望位置的曼哈顿距离
即为：v += abs(i - pos[map[i][j] - 1][0]); v += abs(j - pos[map[i][j] - 1][1]); 大致估算为几十步内得出结果。
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <climits>//形如INT_MAX一类的
#define MAX 100005
#define INF 0x7FFFFFFF
#define REP(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define L(x) x<<1
#define R(x) x<<1|1
# define eps 1e-5
//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂
using namespace std;
int pos[9][2] = { //各个数字的初始位置
{0,0},{0,1},{0,2},
{1,0},{1,1},{1,2},
{2,0},{2,1},{2,2}
};
int map[4][4];
int buff[50];
char input[11];
int limit,ok;
int dx[] = {-1,0,1,0}; //u r d l____0 1 2 3
int dy[] = {0,1,0,-1};
char op[] = {'u','r','d','l'};
int h(int x,int y) {
int v = 0;
for(int i=0; i<3; i++) {
for(int j=0; j<3; j++) {
if(i != x || j != y) {
v += abs(i - pos[map[i][j] - 1][0]);
v += abs(j - pos[map[i][j] - 1][1]);
}
}
}
return v;
}
int dfs(int x,int y,int step,int pre) {
int hn = h(x,y);
if(hn == 0) {
ok = 1;
return step;
}
if(hn + step > limit) return hn + step;
int minn = INF;
for(int i=0; i<4; i++) {
if(abs(i - pre) == 2) continue;
int xx = x + dx[i];
int yy = y + dy[i];
if(xx<0 || xx >=3 || yy<0 || yy >=3) continue;
buff[step] = i;
swap(map[x][y],map[xx][yy]);
int tmp = dfs(xx,yy,step+1,i);
if(ok) return tmp;
minn = min(tmp,minn);
swap(map[x][y],map[xx][yy]);
}
return minn;
}
void IDA_star(int x,int y) {
ok = 0;
memset(buff,-1,sizeof(buff));
while(ok == 0 && limit <= 36) {
limit = dfs(x,y,0,-111);
}
if(ok == 1) {
for(int i=0; i<limit; i++) printf("%c",op[buff[i]]);
puts("");
} else puts("unsolvable");
}
int main() {
for(int i=0; i<9; i++) cin >> input[i];
int t = 0,x,y;
for(int i=0; i<3; i++) {
for(int j=0; j<3; j++) {
if(input[t] == 'x') {
map[i][j] = 9;
x = i;
y = j;
t++;
} else map[i][j] = input[t++] - '0';
}
}
int limit = h(x,y);
if(limit == 0) {
puts("");
return 0;
}
IDA_star(x,y);
return 0;
}
想吐槽一下杭电的这题.........估计各种无法到达目标的数据,所以在输入时候通过求逆序数对数来判断是否有解，本来是TLE，一下蹦到171ms
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <climits>//形如INT_MAX一类的
#define MAX 100005
#define INF 0x7FFFFFFF
#define REP(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define L(x) x<<1
#define R(x) x<<1|1
# define eps 1e-5
//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂
using namespace std;
int pos[9][2] = { //各个数字的初始位置
{0,0},{0,1},{0,2},
{1,0},{1,1},{1,2},
{2,0},{2,1},{2,2}
};
int map[4][4];
int buff[50];
char input[11];
int limit,ok;
int dx[] = {-1,0,1,0}; //u r d l____0 1 2 3
int dy[] = {0,1,0,-1};
char op[] = {'u','r','d','l'};
int h(int x,int y) {
int v = 0;
for(int i=0; i<3; i++) {
for(int j=0; j<3; j++) {
if(i != x || j != y) {
v += abs(i - pos[map[i][j] - 1][0]);
v += abs(j - pos[map[i][j] - 1][1]);
}
}
}
return v;
}
int dfs(int x,int y,int step,int pre) {
int hn = h(x,y);
if(hn == 0) {
ok = 1;
return step;
}
if(hn + step > limit) return hn + step;
int minn = INF;
for(int i=0; i<4; i++) {
if(abs(i - pre) == 2) continue;
int xx = x + dx[i];
int yy = y + dy[i];
if(xx<0 || xx >=3 || yy<0 || yy >=3) continue;
buff[step] = i;
swap(map[x][y],map[xx][yy]);
int tmp = dfs(xx,yy,step+1,i);
if(ok) return tmp;
minn = min(tmp,minn);
swap(map[x][y],map[xx][yy]);
}
return minn;
}
int canget(int a[4][4]) { //这一步省了好多时间 求逆序数对数
int i,j,sum=0,b[10],k=0;
for(i=0; i<3; i++) {
for(j=0; j<3; j++) {
if(a[i][j] != 9)
b[++k]=a[i][j];
}
}
for(i=1; i<=8; i++) {
for(j=1; j<i; j++) {
if(b[i]<b[j])sum++;
}
}
if(sum % 2 ==0)return 1;
else return 0;
}
void IDA_star(int x,int y) {
ok = 0;
memset(buff,-1,sizeof(buff));
while(ok == 0 && limit <= 30) {
limit = dfs(x,y,0,-111);
}
if(ok == 1) {
for(int i=0; i<limit; i++) printf("%c",op[buff[i]]);
puts("");
} else puts("unsolvable");
}
int main() {
while(cin >> input[0]) {
//memset(map,0,sizeof(map));
for(int i=1; i<9; i++) cin >> input[i];
int t = 0,x,y;
for(int i=0; i<3; i++) {
for(int j=0; j<3; j++) {
if(input[t] == 'x') {
map[i][j] = 9;
x = i;
y = j;
t++;
} else map[i][j] = input[t++] - '0';
}
}
limit = h(x,y);
if(limit == 0) {
puts("");
continue;
}
if(canget(map))
IDA_star(x,y);
else puts("unsolvable");
}
return 0;
}
经过猥琐的测试，limit最小限制在29步....................