问题 E: 【排序】绝境求生
时间限制: 1 Sec 内存限制: 64 MB提交: 4 解决: 4
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题目描述
The Eight Puzzle, among other sliding-tile puzzles, is one of the famous problems in artificial intelligence. Along with chess, tic-tac-toe and backgammon, it has been used to study search algorithms.
The Eight Puzzle can be generalized into an M × N Puzzle where at least one of M and N is odd. The puzzle is constructed with MN − 1 sliding tiles with each a number from 1 to MN − 1 on it packed into a M by N frame with one tile missing. For example, with M = 4 and N = 3, a puzzle may look like:
Let's call missing tile 0. The only legal operation is to exchange 0 and the tile with which it shares an edge. The goal of the puzzle is to find a sequence of legal operations that makes it look like:
The following steps solve the puzzle given above.
Given an M × N puzzle, you are to determine whether it can be solved.
The Eight Puzzle can be generalized into an M × N Puzzle where at least one of M and N is odd. The puzzle is constructed with MN − 1 sliding tiles with each a number from 1 to MN − 1 on it packed into a M by N frame with one tile missing. For example, with M = 4 and N = 3, a puzzle may look like:
Let's call missing tile 0. The only legal operation is to exchange 0 and the tile with which it shares an edge. The goal of the puzzle is to find a sequence of legal operations that makes it look like:
The following steps solve the puzzle given above.
Given an M × N puzzle, you are to determine whether it can be solved.
样例输入
3 3
1 0 3
4 2 5
7 8 6
4 3
1 2 5
4 6 9
11 8 10
3 7 0
0 0
样例输出
YES
NO
分析：不知道之前什么时候水过这题，去POJ抄来了题解。。。一个树状数组求逆序对的题。。
#include <iostream> #include <string> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <deque> #include <map> #define range(i,a,b) for(int i=a;i<=b;++i) #define LL long long #define rerange(i,a,b) for(int i=a;i>=b;--i) #define fill(arr,tmp) memset(arr,tmp,sizeof(arr)) using namespace std; int n,m,tol,cnt,head[1000005],ans[1000005],tree[1000005],aa[1000005]; void init(){ } int sum(int x){ int sum=0; while(x>0){ sum+=tree[x]; x-=x&(-x); } return sum; } void add(int x,int y){ while(x<=1000005){ tree[x]+=y; x+=x&(-x); } } void solve(){ while(cin>>n>>m,n,m){ int x,y,t,s=0,num=0; range(i,1,n)range(j,1,m){ cin>>t; if(!t)x=i,y=j; else aa[num++]=t; } fill(tree,0); rerange(i,num-1,0){ s+=sum(aa[i]-1); add(aa[i],1); } if(m&1)cout<<(s&1?"NO":"YES")<<endl; else if(((n-x)^s)&1)cout<<"NO"<<endl; else cout<<"YES"<<endl; } } int main() { init(); solve(); return 0; }