【題目鏈接】:http://hihocoder.com/problemset/problem/1312?sid=1092363

【題意】

【題解】

定義一個(gè)A*函數(shù)
f = step+val
這里的val是當(dāng)前這個(gè)狀態(tài);每個(gè)點(diǎn)到目標(biāo)狀態(tài)的點(diǎn)的曼哈頓距離的絕對(duì)值;
(這個(gè)值肯定比真正需要花費(fèi)的路程短)
step就為當(dāng)前狀態(tài)花費(fèi)的步數(shù);
把普通隊(duì)列改成優(yōu)先隊(duì)列;
優(yōu)先處理f值小的狀態(tài);
f值相同的,優(yōu)先處理step值小的;
(也就是說f值大的不是不處理了,而是放到后面再處理)
這樣就能較快地逼近目標(biāo)狀態(tài)了;
效果異常地棒
2s變成0.2s了!

【Number Of WA】

0

【完整代碼】

#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; }

轉(zhuǎn)載于:https://www.cnblogs.com/AWCXV/p/7626359.html

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