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1105. Spiral Matrix (25) -- 模拟

2017-02-14 19:18 417 查看

题目地址

https://www.patest.cn/contests/pat-a-practise/1105

ac代码

思路是:外围转一圈的函数封装起来,然后遍历填充,注意判断是否完成了

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
#include <sstream>
#include <list>
#include <stack>
#include <map>
#include <set>
#include <iterator>
#include <unordered_map>

using namespace std;

#define INF 0x7fffffff

typedef long long int LL;

const int N = 404;

int n;
vector<int> v;

bool cmp(int a, int b)
{
return a > b;
}

int r;
int c;

vector<vector<int>> ans;

void func(int stai,int staj,int endi,int endj,int index)
{
if(stai == endi) // 只有一行
{
for(int j=staj;j<=endj;j++){
ans[stai][j] = v[index++];
if(index > n)
break;
}
return;
}
if(staj == endj) // 只有一列
{
for(int i=stai;i<=endi;i++){
ans[i][staj] = v[index++];
if(index > n)
break;
}
return;
}

for(int j=staj;j<=endj;j++){
ans[stai][j] = v[index++];
if(index > n)
break;
}
for(int i = stai + 1; i <= endi;i++)
{
ans[i][endj] = v[index++];
if(index > n)
break;
}
for(int j = endj - 1; j >= staj ; j--)
{
ans[endi][j] = v[index++];
if(index > n)
break;
}
for(int i= endi - 1;i>=stai+1;i--)
{
ans[i][staj] = v[index++];
if(index > n)
break;
}
}

int main()
{
//freopen("in.txt", "r" , stdin);

while(scanf("%d", &n) != EOF)
{
for(int i=0;i<n;i++)
{
int tmp;
scanf("%d", &tmp);
v.push_back(tmp);
}

sort(v.begin(), v.end(), cmp);

r = n;
c = 1;
int cha = r - c;

for(int i=1;i<=n;i++)
{
for(int j=1;j<=i;j++)
{
if(i * j == n)
{
if(i - j < cha)
{
r = i;
c = j;
cha = i - j;
}
}
}
}

vector<vector<int>> ansT(r, vector<int>(c));
ans = ansT;

int stai = 0;
int endi = r - 1;
int staj = 0;
int endj = c - 1;
int index = 0;
while(index < n)
{
int cnt = 2 * (endj - staj + 1) + 2 * (endi - stai + 1 - 2);
func(stai, staj, endi, endj, index);
index += cnt;
stai ++;
staj ++;
endi --;
endj --;
}
for(int i=0;i<r;i++)
{
cout << ans[i][0];
for(int j=1;j<c;j++)
{

cout << " " << ans[i][j];
}
cout << endl;
}
}
return 0;
}
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