题目链接：http://poj.org/problem?id=4012
Problem B. Black Square
Input ?le: black.in
Output ?le: black.out
Time limit: 3 seconds
Memory limit: 256 megabytes
Inspired by Kazimir Malevich’s masterpiece “Black Square”, Peter Palevich is planning to create his own version. He prepared a rectangular grid containing m×n white cells arranged in m rows of n cells each. Peter painted some of the cells black, so that the
black cells formed a square of size s×s cells. But later that day Peter became disappointed with his painting and destroyed it, cutting it to horizontal stripes of size 1×n and burning them in the ?replace. Next morning Peter changed his mind and decided to
restore his painting. He tried to ?nd its remains in the ?replace, and fortunately one of the stripes, namely the k-th from the top, survived the ?re. Now Peter wonders whether it is possible to restore the painting based on this stripe. Help him to do it.
Input The ?rst line of the input ?le contains four integer numbers: m, n, s and k (1 ≤ m,n ≤ 5000; 1≤ s ≤min(m,n); 1≤ k ≤ m). The second line contains n characters and describes the k-th line of the painting, ‘.’ stands for a white cell, ‘*’ stands for a black
cell. Output If the initial painting can be uniquely restored, output “Unique”. If there are several paintings that could have been painted by Peter, output “Ambiguous”. If there are no possible paintings, output “Impossible”.
Examples
black.in black.out
4 4 1 2 ..*.
Unique
4 4 2 2 ..**
Ambiguous
4 4 3 2 .*.*
Impossible
分类讨论：
首先统计出所给行中black的个数，如果中间有两个black之间有white，则必然impossible，不用再判断了
否则分为以下几种情况：
1. black的个数 == 0：这时说明所给的行为空行，需要判断空行以上和以下能否放下 s*s 的cell，判断是否唯一，多种情况，或者不可能（代码中的judge（））；
2. black的个数 ！= s，此时为impossible；
3. blackd个数 == s， 只有一个black块 或者 所给行为第一行 或者 所给行是最后一行 或者 s==m
都是唯一的
4.否则，都是多解
晕~~~~
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int m,n,s,k,flag,black;
int judge(int x)
{
if(x < s|| n < s) return 0;
if(x == s && n== s) return 1;
return 2;
}
int main()
{
char a[5005];
while(scanf("%d%d%d%d",&m,&n,&s,&k) !=EOF)
{
scanf("%s",a);
flag = 0;
black = 0;
for(int i = 0;i < strlen(a);i ++)
{
if(flag == 0)
{
if(a[i] == '*')
{
black ++;
flag = 1;
}
}
else if(flag == 1)
{
if(a[i] == '*')
black++;
else
flag = 2;
}
else if(flag == 2)
{
if(a[i] == '*')
black = -1;
}
}
int up = k-1,down = m-k;
int ans;
if(black == 0)
ans = judge(up) + judge(down);
else if(black != s)
ans = 0;
else if( s==1 || m==s || k==1 || k==m)
ans = 1;
else
ans = 2;
if(ans == 1)
printf("Unique\n");
else if(ans == 0)
printf("Impossible\n");
else
printf("Ambiguous\n");
}
return 0;
}
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