【题目链接】:http://hihocoder.com/problemset/problem/1312?sid=1092363
【题意】
【题解】
定义一个A*函数
f = step+val
这里的val是当前这个状态;每个点到目标状态的点的曼哈顿距离的绝对值;
(这个值肯定比真正需要花费的路程短)
step就为当前状态花费的步数;
把普通队列改成优先队列;
优先处理f值小的状态;
f值相同的,优先处理step值小的;
(也就是说f值大的不是不处理了,而是放到后面再处理)
这样就能较快地逼近目标状态了;
效果异常地棒
2s变成0.2s了!
【Number Of WA】
0
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)typedef pair<int,int> pii;
typedef pair<LL,LL> pll;const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const int pre[9][2] =
{{2,2},{0,0},{0,1},{0,2},{1,0},{1,1},{1,2},{2,0},{2,1}
};
const double pi = acos(-1.0);
const int N = 110;struct node
{int a[9],p,step,f;friend bool operator < (node x,node y){if (x.f==y.f){if (x.step==y.step)return true;elsereturn x.step>y.step;}elsereturn x.f>y.f;}
};node init;
priority_queue <node> dl;
map <int,int> dic;
int a[9],goal;
int cs[9] = {1,2,3,4,5,6,7,8,0};int has(int *a)
{int x = 0;rep1(i,0,8) x = x*10 + a[i];return x;
}int val(int *a)
{int ret = 0;rep1(i,0,8){int x = i/3,y = i%3;if (a[i]==0) continue;ret+=abs(pre[a[i]][0]-x)+abs(pre[a[i]][1]-y);}return ret;
}int bfs()
{while (!dl.empty()){int p = dl.top().p;int now = dl.top().step;node temp;rep1(i,0,8) temp.a[i] = dl.top().a[i];dl.pop();//上if (p>2){int tp = p-3;swap(temp.a[tp],temp.a[p]);int xzt = has(temp.a);if (xzt==goal) return now+1;if (dic.find(xzt)==dic.end()){dic[xzt] = now+1;temp.step = now+1;temp.p = tp;temp.f = temp.step+val(temp.a);dl.push(temp);}swap(temp.a[tp],temp.a[p]);}//下if (p<6){int tp = p+3;swap(temp.a[tp],temp.a[p]);int xzt = has(temp.a);if (xzt==goal) return now+1;if (dic.find(xzt)==dic.end()){dic[xzt] = now+1;temp.step = now+1;temp.p = tp;temp.f = temp.step+val(temp.a);dl.push(temp);}swap(temp.a[tp],temp.a[p]);}//左if (p%3!=0){int tp = p-1;swap(temp.a[tp],temp.a[p]);int xzt = has(temp.a);if (xzt==goal) return now+1;if (dic.find(xzt)==dic.end()){dic[xzt] = now+1;temp.step = now+1;temp.p = tp;temp.f = temp.step+val(temp.a);dl.push(temp);}swap(temp.a[tp],temp.a[p]);}//右if (p%3!=2){int tp = p+1;swap(temp.a[tp],temp.a[p]);int xzt = has(temp.a);if (xzt==goal) return now+1;if (dic.find(xzt)==dic.end()){dic[xzt] = now+1;temp.step = now+1;temp.p = tp;temp.f = temp.step+val(temp.a);dl.push(temp);}swap(temp.a[tp],temp.a[p]);}}return -1;
}int main()
{//freopen("F:\\rush.txt","r",stdin);ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not usegoal = has(cs);int t;cin >> t;while (t--){dic.clear();rep1(i,0,8){cin >> a[i];if (a[i]==0) init.p = i;}rep1(i,0,8) init.a[i] = a[i];init.step = 0,init.f = init.step+val(init.a);dic[has(init.a)] = 0;if (has(init.a)==goal){cout << 0 << endl;continue;}while (!dl.empty()) dl.pop();dl.push(init);int ans = bfs();if (ans==-1)cout <<"No Solution!"<<endl;elsecout << ans << endl;}return 0;
}
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