http://www.lydsy.com/JudgeOnline/problem.php?id=2102

直接枚举所有情况。。。。。。然后判断是否可行。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr2(a, b, c) for1(i, 1, b) { for1(j, 1, c) cout << a[i][j]; cout << endl; }
#define printarr1(a, b) for1(i, 1, b) cout << a[i]; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; } const int N=105;
int n, m, cnt, ans, a[N][N], b[N]; bool check(int t) {
for1(i, 1, m) {
int tot=0;
rep(j, n) if(a[i][j] && (t&(1<<j))) ++tot;
if(tot!=b[i]) return 0;
}
return 1;
} int main() {
read(n); read(m);
for1(i, 1, m) {
rep(j, n) {
char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
a[i][j]=ch-'0';
}
b[i]=getint();
}
int mx=(1<<n)-1, flag=0;
for1(x, 0, mx) {
if(check(x)) {
if(cnt) { puts("NOT UNIQUE"); return 0; }
flag=1;
ans=x;
++cnt;
}
}
if(!flag) puts("IMPOSSIBLE");
else rep(i, n) printf("%d", (bool)(ans&(1<<i)));
return 0;
}

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

Sample Output

1010

HINT

Source

【BZOJ】2102: [Usaco2010 Dec]The Trough Game(暴力)的更多相关文章

  1. 2102: [Usaco2010 Dec]The Trough Game

    2102: [Usaco2010 Dec]The Trough Game Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 117  Solved: 84[ ...

  2. BZOJ 2100: [Usaco2010 Dec]Apple Delivery( 最短路 )

    跑两遍最短路就好了.. 话说这翻译2333 ---------------------------------------------------------------------- #includ ...

  3. BZOJ 2101: [Usaco2010 Dec]Treasure Chest 藏宝箱( dp )

    dp( l , r ) = sum( l , r ) - min( dp( l + 1 , r ) , dp( l , r - 1 ) ) 被卡空间....我们可以发现 l > r 是无意义的 ...

  4. BZOJ2102 : [Usaco2010 Dec]The Trough Game

    暴力枚举答案然后检验. #include<cstdio> int n,m,i,j,k,a[100],b[100],cnt,ans;char s[20]; int main(){ for(s ...

  5. BZOJ 2101 [Usaco2010 Dec]Treasure Chest 藏宝箱:区间dp 博弈【两种表示方法】【压维】

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=2101 题意: 共有n枚金币,第i枚金币的价值是w[i]. 把金币排成一条直线,Bessie ...

  6. BZOJ——2101: [Usaco2010 Dec]Treasure Chest 藏宝箱

    http://www.lydsy.com/JudgeOnline/problem.php?id=2101 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit:  ...

  7. bzoj 2097: [Usaco2010 Dec]Exercise 奶牛健美操【二分+树形dp】

    二分答案,然后dp判断是否合法 具体方法是设f[u]为u点到其子树中的最长链,每次把所有儿子的f值取出来排序,如果某两条能组合出大于mid的链就断掉f较大的一条 a是全局数组!!所以要先dfs完子树才 ...

  8. bzoj 2101: [Usaco2010 Dec]Treasure Chest 藏宝箱【区间dp】

    就是区间dp啦f[i][j]表示以i开头的长为j+1的一段的答案,转移是f[i][j]=s[i+l]-s[i-1]+min(f[i][j-1],f[i+1][j-1]),初始是f[i][1]=a[i] ...

  9. bzoj 2100: [Usaco2010 Dec]Apple Delivery【spfa】

    洛谷数据好强啊,普通spfa开o2都过不了,要加双端队列优化 因为是双向边,所以dis(u,v)=dis(v,u),所以分别以pa1和pa2为起点spfa一遍,表示pb-->pa1-->p ...

随机推荐

  1. XP如何找到网上邻居

    右击桌面,点击属性,切换到桌面,自定义桌面,勾选网上邻居即可.

  2. linux中添加开机自启服务的方法

    往文件/etc/rc.d/rc.local中追加内容即可,如: /mongodb/start_mongoDB.sh

  3. 用filter:grayscale将图片过滤成灰色

    设置成百分之百直接过滤成灰色: img{filter:gray; filter:grayscale(100%); -0-filter:grayscale(100%); -moz-filter:gray ...

  4. 阻塞与非阻塞、同步与异步、I/O模型

    1. 概念理解 在进行网络编程时,我们常常见到同步(Sync)/异步(Async),阻塞(Block)/非阻塞(Unblock)四种调用方式: 同步/异步主要针对C端:  同步: 所谓同步,就是在c端 ...

  5. mac 查看目前哪些进程占用哪些端口

    lsof -nP  | grep TCP | grep LISTEN lsof -i :TCP

  6. jQuery Growl 插件(消息提醒)

    jQuery Growl 插件(消息提醒) 允许您很容易地在一个覆盖层显示反馈消息.消息会在一段时间后自动消失,不需要单击"确定"按钮等.用户也可以通过移动鼠标或点击关闭按钮加快隐 ...

  7. QT Unexpected CDB exit 问题的解决办法

    行QT进行debug时,提示 Unexpected CDB exit ,The CBD process terminated.. QtCreator 默认是没有调试器的,因此需要用户额外安装. win ...

  8. python selenium --处理下拉框

    下拉框是我们最常见的一种页面元素,对于一般的元素,我们只需要一次就定位,但下拉框里的内容需要进行两次定位,先定位到下拉框,再定位到下拉框内里的选项. drop_down.html <html&g ...

  9. atitit.  web组件化原理与设计

    atitit.  web组件化原理与设计 1. Web Components提供了一种组件化的推荐方式,具体来说,就是:1 2. 组件化的本质目的并不一定是要为了可复用,而是提升可维护性. 不具有复用 ...

  10. 新型I/O架构引领存储之变(二)

    新型I/O架构引领存储之变(二) 作者:廖恒 众所周知,支持存储及网络I/O服务的接口协议有很多种.比方,以太网及Infiniband接口都支持採用iSCSI协议来实现存储业务,它们也因而成为了ser ...