codeforces 26C Parquet 贪心 模拟

Parquet

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

Once Bob decided to lay a parquet floor in his living room. The living room is of size n × m metres. Bob had planks of three types: a planks 1 × 2 meters, b planks 2 × 1 meters, and c planks 2 × 2 meters. Help Bob find out, if it is possible to parquet the living room with such a set of planks, and if it is possible, find one of the possible ways to do so. Bob doesn’t have to use all the planks.

Input

The first input line contains 5 space-separated integer numbers n, m, a, b, c (1 ≤ n, m ≤ 100, 0 ≤ a, b, c ≤ 104), n and m — the living room dimensions, a, b and c — amount of planks 1 × 2, 2 × 1 и 2 × 2 respectively. It’s not allowed to turn the planks.

Output

If it is not possible to parquet the room with such a set of planks, output IMPOSSIBLE. Otherwise output one of the possible ways to parquet the room — output n lines with m lower-case Latin letters each. Two squares with common sides should contain the same letters, if they belong to one and the same plank, and different letters otherwise. Different planks can be marked with one and the same letter (see examples). If the answer is not unique, output any.

Examples

input

2 6 2 2 1

output

aabcca

aabdda

input

1 1 100 100 100

output

IMPOSSIBLE

input

4 4 10 10 10

output

aabb

aabb

bbaa

bbaa

题目链接

题意：你有1*2,2*1和2*2的木板各a,b,c个，然后问你能否用这些木板拼成n*m的木板。

解题思路：明显的贪心题，刚开始却不知道从何下手，可以知道的是我们肯定可以先把2*2的木板全部用完，因为没有想过可以把2*2的木板先取负，后来才想到可以，然后用余下的1*2和2*1的木板对其进行补足就好了。如果木板不足或是n*m无法整除2，那么我们就无法拼成n*m的木板。

然后之后还需要对其进行编号，因为是100*100，所以可以直接暴力找编号就好，就是比较的时候真的是烦到不行…

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,a,b,c,st[105][105],flag=1;
char st1[105][105];
void cc(int i,int j){
st[i][j]=st[i+1][j]=st[i][j+1]=st[i+1][j+1]=3;
}
void bb(int i,int j){
st[i][j]=st[i+1][j]=2;
}
void aa(int i,int j){
st[i][j]=st[i][j+1]=1;
}

void aaa(int x,int y){
char s;
for(char i='a';;i++){
if(i!=st1[x-1][y]&&i!=st1[x+1][y]&&i!=st1[x][y-1]
&&i!=st1[x][y+2]&&i!=st1[x-1][y+1]&&i!=st1[x+1][y+1]){
s=i;
break;
}
}
st1[x][y]=st1[x][y+1]=s;
st[x][y]=st[x][y+1]=0;
}

void bbb(int x,int y){
char s;
for(char i='a';;i++){
if(i!=st1[x-1][y]&&i!=st1[x+2][y]&&i!=st1[x][y-1]
&&i!=st1[x][y+1]&&i!=st1[x+1][y+1]&&i!=st1[x+1][y-1]){
s=i;
break;
}
}
st1[x][y]=st1[x+1][y]=s;
st[x][y]=st[x+1][y]=0;
}

void ccc(int x,int y){
char s;
for(char i='a';;i++){
if(i!=st1[x-1][y]&&i!=st1[x-1][y+1]&&i!=st1[x][y-1]
&&i!=st1[x][y+2]&&i!=st1[x+1][y-1]&&i!=st1[x+1][y+2]
&&i!=st1[x+2][y]&&i!=st1[x+2][y+1]){
s=i;
break;
}
}
st1[x][
4000
y]=st1[x][y+1]=st1[x+1][y]=st1[x+1][y+1]=s;
st[x][y]=st[x][y+1]=st[x+1][y]=st[x+1][y+1]=0;
}
int main(){
scanf("%d%d%d%d%d",&n,&m,&a,&b,&c);
if((m*n)%2==1||m*n>2*a+2*b+4*c){
printf("IMPOSSIBLE\n");
return 0;
}
c-=(m/2)*(n/2);
for(int i=1;i<=n;i+=2){
for(int j=1;j<=m;j+=2){
cc(i,j);
}
}
if(m%2==1){
for(int i=1;i<=n;i+=2){
bb(i,m);
b--;
}
}
if(n%2==1){
for(int i=1;i<=m;i+=2){
aa(n,i);
a--;
}
}
if(a<0||b<0){
printf("IMPOSSIBLE\n");
return 0;
}
if(c<0){
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(st[i][j]==3){
if(a>=2){
aa(i,j);
aa(i+1,j);
a-=2;
c++;
}
else if(b>=2){
bb(i,j);
bb(i,j+1);
b-=2;
c++;
}
else{
flag=0;
break;
}
}
if(c==0)    break;
}
if(flag==0) break;
if(c==0)    break;
}
}
if(flag==0){
printf("IMPOSSIBLE\n");
return 0;
}
else{
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(st[i][j]==0) continue;
else if(st[i][j]==1){
aaa(i,j);
}
else if(st[i][j]==2){
bbb(i,j);
}
else if(st[i][j]==3){
ccc(i,j);
}
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++)   printf("%c",st1[i][j]);
printf("\n");
}
}
return 0;
}