「BZOJ2102」[Usaco2010 Dec] The Trough Game
Description
Farmer John and Bessie are playing games again. This one has to do with troughs of water. Farmer John has hidden N (1 <= N <= 20) troughs behind the barn, and has filled some of them with food. Bessie has asked M (1 <= M <= 100) questions of the form, “How many troughs from this list (which she recites) are filled?”. Bessie needs your help to deduce which troughs are actually filled. Consider an example with four troughs where Bessie has asked these questions (and received the indicated answers): 1) “How many of these troughs are filled: trough 1” –> 1 trough is filled 2) “How many of these troughs are filled: troughs 2 and 3” –> 1 trough is filled 3) “How many of these troughs are filled: troughs 1 and 4” –> 1 trough is filled 4) “How many of these troughs are filled: troughs 3 and 4” –> 1 trough is filled From question 1, we know trough 1 is filled. From question 3, we then know trough 4 is empty. From question 4, we then know that trough 3 is filled. From question 2, we then know that trough 2 is empty. 求N位二进制数X，使得给定的M个数，满足X and Bi=Ci ,Bi ci分别是读入的两个数
Input
* Line 1: Two space-separated integers: N and M * Lines 2..M+1: A subset of troughs, specified as a sequence of contiguous N 0’s and 1’s, followed by a single integer that is the number of troughs in the specified subset that are filled.
Output
* Line 1: A single line with: * The string “IMPOSSIBLE” if there is no possible set of filled troughs compatible with Farmer John’s answers. * The string “NOT UNIQUE” if Bessie cannot determine from the given data exactly what troughs are filled. * Otherwise, a sequence of contiguous N 0’s and 1’s specifying which troughs are filled.
Sample Input
4 4
1000 1
0110 1
1001 1
0011 1
1000 1
0110 1
1001 1
0011 1
Sample Output
1010
题解
直接暴力枚举
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#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> #include<map> #define inf 1000000000 using namespace std; inline int 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,cnt; int a[105][25],b[105],now[25]; int ans[25]; char ch[20]; bool jud() { for(int i=1;i<=m;i++) { int tot=0; for(int j=1;j<=n;j++) if(a[i][j]&now[j]) tot++; if(tot!=b[i])return 0; } return 1; } void solve() { int t=pow(2,n),tmp; while(t--) { if(jud()) { for(int i=1;i<=n;i++)ans[i]=now[i]; cnt++; if(cnt>=2)return; } now[n]++;tmp=n; while(now[tmp]==2){now[tmp]=0;tmp--;now[tmp]++;} } } int main() { n=read();m=read(); for(int i=1;i<=m;i++) { scanf("%s",ch); for(int j=1;j<=n;j++) a[i][j]=ch[j-1]-'0'; b[i]=read(); } solve(); if(!cnt)puts("IMPOSSIBLE"); else if(cnt==1) for(int i=1;i<=n;i++)printf("%d",ans[i]); else puts("NOT UNIQUE"); return 0; } |
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计数器 9857441
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