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)=60
int 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 squares
s = 0;
memset(contains, 0, sizeof(contains));
for(int i = 1; i <= n; i++) // side length
for(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 square
for(int j = 0; j < i; j++) {
int a = row_match(x, y+j); // up
int b = row_match(x+i, y+j); // down
int c = col_match(x+j, y); // left
int d = col_match(x+j, y+i); // right
contains[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 found
return;
}
// remove a match in that square
for(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