「BZOJ1647」[Usaco2007 Open] Fliptile 翻格子游戏
Description
Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M x N grid (1 <= M <= 15; 1 <= N <= 15) of square tiles, each of which is colored black on one side and white on the other side. As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make. Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word “IMPOSSIBLE”.
Input
* Line 1: Two space-separated integers: M and N
* Lines 2..M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white
Output
* Lines 1..M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.
Sample Input
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1
Sample Output
1 0 0 1
1 0 0 1
0 0 0 0
OUTPUT DETAILS:
After flipping at row 2 column 1, the board will look like:
0 0 0 1
1 0 1 0
1 1 1 0
1 0 0 1
After flipping at row 2 column 4, the board will look like:
0 0 0 0
1 0 0 1
1 1 1 1
1 0 0 1
After flipping at row 3 column 1, the board will look like:
0 0 0 0
0 0 0 1
0 0 1 1
0 0 0 1
After flipping at row 3 column 4, the board will look like:
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
Another solution might be:
0 1 1 0
0 0 0 0
0 0 0 0
0 1 1 0
but this solution is lexicographically higher than the solution above.
题解
唉，枚举第一行的状态，然后可以推出其它格子的状态
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#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #define ll long long #define inf 1000000000 using namespace std; inline ll read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,m,ed,mn=inf; int xx[4]={0,0,1,-1},yy[4]={1,-1,0,0}; bool a[16][16],t[16][16],op[16][16],ans[16][16]; void operate(int x,int y) { op[x][y]=1;t[x][y]^=1; for(int k=0;k<4;k++) { int nowx=x+xx[k],nowy=y+yy[k]; if(nowx<1||nowy<1||nowx>n||nowy>m)continue; t[nowx][nowy]^=1; } } void solve(int x) { memset(op,0,sizeof(op)); int tot=0; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) t[i][j]=a[i][j]; for(int i=1;i<=m;i++) if((1<<(i-1))&x) { tot++;if(tot>=mn)return; operate(1,m-i+1); } for(int i=2;i<=n;i++) for(int j=1;j<=m;j++) { if(t[i-1][j]) { tot++;if(tot>=mn)return; operate(i,j); } } for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(t[i][j])return; mn=tot; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) ans[i][j]=op[i][j]; } int main() { n=read();m=read(); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) a[i][j]=read(); ed=(1<<m)-1; for(int i=0;i<=ed;i++) solve(i); if(mn!=inf) { for(int i=1;i<=n;i++) { for(int j=1;j<m;j++) printf("%d ",ans[i][j]); printf("%d\n",ans[i][m]); } } else puts("IMPOSSIBLE"); return 0; } |
OI 课件=>
黄学长，请问那个cnt和tot是什么作用，怎么判断两个方案字典序大小
我并没有看到cnt，tot是用来记录操作次数的吧
枚举第一行的时候就按照字典序来枚举