题目地址：http://poj.org/problem?id=1084
看到很多用Dancing Link X 算法，但没接触过，而且是为了练习A*，所以就想A*
一开始用 还有几个正方形数目代表h()估价函数，但是答案错了，因为有的火柴能破坏3个，有的2个，有的1 个....不能代表估价函数，并没有相容性
错误代码如下：
#include<iostream>
#include<cstdio>
#include<set>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
const int maxn=60+5;
int n,tot;
struct Node{bool Gird[maxn];int f,g,h;Node(bool *G,int g,int h):g(g),h(h){memcpy(Gird,G,sizeof(bool)*(tot+2));f=g+h;}bool operator < (const Node& n) const {return f>n.f;}
};
int SquareNum(bool *G,int i,int t){ //i为上边 ，一共有几个正方形 int cnt=0;int L,R;bool ok;for(int j=0;j<=t-i;j++){L=(i+n)+(2*n+1)*j,R=i+j;if(G[L]&&G[R]&&L<=tot){L+=n+1,R+=n+1; ok=true;for(int k=0;k<=j&&ok;k++){if(!G[L]||!G[R]||L>tot) ok=false;L+=1,R+=2*n+1; }if(ok) cnt++;}else break;}return cnt;
}
int H(bool *G){int cnt=0;for(int j=n;j<tot;j+=2*n+1) //有几个正方形就是至少需要几个步骤 for(int i=j-n+1;i<=j;i++) cnt+=SquareNum(G,i,j);return cnt;
}
void Switch(bool *Gird,int n){Gird[n]^=true; //取反
}
set<LL> closed;
bool inClosed(bool *Gird){LL code=0;for(int i=1;i<=tot;i++){ //最多60根 所以状态压缩 long long 就能保存 if(Gird[i]) code|=1;code<<=1;}if(closed.count(code)) return true;closed.insert(code);return false;
}
int A_Star(Node s)
{closed.clear();priority_queue<Node> open;open.push(s);inClosed(s.Gird);while(!open.empty()){Node u=open.top(); open.pop();if(u.h==0) return u.g;for(int i=1;i<=tot;i++){if(!u.Gird[i]) continue;bool G[maxn]; memcpy(G,u.Gird,sizeof(G));Switch(G,i);if(inClosed(G)) continue;open.push(Node(G,u.g+1,H(G))); }}
}
int main()
{int T; bool Gird[maxn]; int m;cin>>T;while(T--){cin>>n>>m;tot=2*n*n+2*n; //n*n总共有2*n*(n+1)个火柴 memset(Gird,true,sizeof(Gird));while(m--){int x; cin>>x;Switch(Gird,x);}cout<<A_Star(Node(Gird,0,H(Gird)))<<endl;}return 0;
}一直在思考估价函数怎么表示...
看了刘汝佳大神算法，BFS用最优性剪枝就搞定了
思路如下：以每一个正方形为对象，并且必须要以从小正方形到大的顺序，dfs拿去正方形的一条边
而且判断正方形的算法特别巧妙，要事先给原图存在的正方形全部标号，给每个正方形标记存在的边和应该存在边，如1号小正方形存在3条边，而应该存在4条边，所以1号无正方形
代码如下：
// UVa1603 Square Destroyer
// Rujia Liu
// This code implements a variant of an algorithm presented in a book. It's simple yet efficient.
// Readers are encouraged to experiment on other algorithms.
// However, it's still slow for n=5 and m=0 (which is NOT in judge input)
// If you really want an efficient solution, learn DLX (Algorithm X with dancing links)
// DLX is well expained (with code) in my other book <<Beginning Algorithm Contests -- Training Guide>>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;const int maxs = 60; // number of squares: 25+16+9+4+1=55
const int maxm = 60; // number of matches: 2*5*(5+1)=60int n, exists[maxm]; // matches 是否有这个火柴
int s, size[maxs], fullsize[maxs], contains[maxs][maxm]; // squares
int best; //给矩形编号，size[s]指第s个矩形实际有几条边存在，fullsize[s]指s矩形应该存在条边
//best是答案 // contain[s][e] if true ,代表s矩形含有e这条边 inline int row_match(int x, int y) {return (2*n+1)*x+y;
}inline int col_match(int x, int y) {return (2*n+1)*x+n+y;
}// number of matches in a full n*n grid
inline int match_count(int n) {return 2*n*(n+1);
}void init() {int m, v;scanf("%d%d", &n, &m);for(int i = 0; i < match_count(n); ++i) exists[i] = 1;while(m--) {scanf("%d", &v);exists[v-1] = 0;}// collect full squaress = 0;memset(contains, 0, sizeof(contains));for(int i = 1; i <= n; i++) // side lengthfor(int x = 0; x <= n-i; x++)for(int y = 0; y <= n-i; y++) {size[s] = 0;fullsize[s] = 4*i; // number of matches in a complete squarefor(int j = 0; j < i; j++) {int a = row_match(x, y+j); // upint b = row_match(x+i, y+j); // downint c = col_match(x+j, y); // leftint d = col_match(x+j, y+i); // rightcontains[s][a] = 1;contains[s][b] = 1;contains[s][c] = 1;contains[s][d] = 1;size[s] += exists[a] + exists[b] + exists[c] + exists[d]; // number of matches now}++s;}
}int find_square() {for(int i = 0; i < s; i++)if(size[i] == fullsize[i]) return i;return -1;
}void dfs(int dep) {if(dep >= best) return; //最优性剪枝 int k = find_square(); //随意找个矩形(完整的) if(k == -1) {best = dep; //404: not foundreturn;}// remove a match in that squarefor(int i = 0; i < match_count(n); i++)if(contains[k][i]) {for(int j = 0; j < s; j++)if(contains[j][i]) size[j]--; //delete dfs(dep + 1);for(int j = 0; j < s; j++) //restore if(contains[j][i]) size[j]++;}
}int main() {int T;scanf("%d", &T);while(T--) {init();best = n*n;dfs(0);printf("%d\n", best);}return 0;
}Dancing Link 以后学好再写