USACO 2.2 Party Lamps
2015-04-30 12:54
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这道题的闲话时间我一定要放在开头……
觉得自己这道题实在写得太丑了,但是最近事情多,刷过了懒得改,于是我决定随便说一说自己的思路和官方的思路,然后再把自己丑得要命的代码贴一下……
一开始我的思路是,1234四种按钮,它们影响的灯各不相同,总而言之,根据灯的编号就能明确哪几个按钮可以影响到它,和按钮数应该是单数还是双数。将这个整理出来。
由于1234这四种按钮按的数量不是单数就是双数,所以遍历的话也就是2^4的数量。遍历从中取出满足条件的。
就这样……虽然听上去好像不太复杂,实际上这道题我用了超过一百行的代码,看上去错综复杂,简直了……
刷完题之后我很快地看了官方解,它的第一个思路就呼我一脸……这道题虽然看上去灯数很多,但是说到底就是6个灯的事儿,只要研究好一组6个灯的情况,就可以输出整体的情况。判断的方法应该跟我原来的思路差不太多,但是代码上清晰很多。
以下就是我的代码:
#include <iostream>
#include <fstream>
using namespace std;
const int MAXN = 101;
int s24 = 0;
int s3 = 0;
int s2 = 0;
int s34 = 0;
int N, C;
int res[16][MAXN] = {0};
int resLen = 0;
bool omits[16] = {0};
bool initData()
{
ifstream fin ("lamps.in");
fin >> N >> C;
int t;
for (fin >> t; t != -1; fin >> t) {
if (t % 6 == 1)
s24 = 2;
else if (t % 6 == 4)
s34 = 2;
else if (t % 2 == 0)
s3 = 2;
else
s2 = 2;
}
for (fin >> t; t != -1; fin >> t) {
if (t % 6 == 1) {
if (s24 != 2)
s24 = 1;
else
return false;
}
else if (t % 6 == 4) {
if (s34 != 2)
s34 = 1;
else
return false;
}
else if (t % 2 == 0) {
if (s3 != 2)
s3 = 1;
else
return false;
}
else {
if (s2 != 2)
s2 = 1;
else
return false;
}
}
return true;
}
void findCombos()
{
int i, j, k, m, n;
for (i = 0; i < 2; i ++) {
for (j = 0; j < 2; j ++) {
for (k = 0; k < 2; k ++) {
for (m = 0; m < 2; m ++) {
if (i+j+k+m > C)
continue;
if ((s24 && (s24+i+j+m)%2)
|| (s3 && (s3+i+k)%2)
|| (s2 && (s2+i+j)%2)
|| (s34 && (s34+i+k+m)%2))
continue;
for (n = 1; n <= N; n ++)
res[resLen]
= (i + (n%2?j:k) + (n%3==1?m:0) + 1) % 2;
resLen ++;
}
}
}
}
}
void sortCombos()
{
int i, j, k, t;
bool isS;
for (i = 0; i < resLen - 1; i ++) {
for (j = 0; j < resLen - i - 1; j ++) {
isS = false;
for (k = 1; k <= N; k ++) {
if (res[j][k] < res[j+1][k])
break;
else if (res[j][k] > res[j+1][k]) {
isS = true;
break;
}
}
if (isS) {
for (k = 1; k <= N; k ++) {
t = res[j][k];
res[j][k] = res[j+1][k];
res[j+1][k] = t;
}
}
}
}
for (i = 0; i < resLen - 1; i ++) {
isS = true;
for (j = 1; j <= N; j ++) {
if (res[i][j] != res[i+1][j]) {
isS = false;
break;
}
}
if (isS)
omits[i] = 1;
}
}
int main()
{
ofstream fout ("lamps.out");
bool f;
f = initData();
if (!f) {
fout << "IMPOSSIBLE" << endl;
return 0;
}
findCombos();
sortCombos();
if (resLen == 0) {
fout << "IMPOSSIBLE" << endl;
return 0;
}
int i, j, k;
for (i = 0; i < resLen; i ++) {
if (!omits[i]) {
for (j = 1; j <= N; j ++)
fout << res[i][j];
fout << endl;
}
}
}
觉得自己这道题实在写得太丑了,但是最近事情多,刷过了懒得改,于是我决定随便说一说自己的思路和官方的思路,然后再把自己丑得要命的代码贴一下……
一开始我的思路是,1234四种按钮,它们影响的灯各不相同,总而言之,根据灯的编号就能明确哪几个按钮可以影响到它,和按钮数应该是单数还是双数。将这个整理出来。
由于1234这四种按钮按的数量不是单数就是双数,所以遍历的话也就是2^4的数量。遍历从中取出满足条件的。
就这样……虽然听上去好像不太复杂,实际上这道题我用了超过一百行的代码,看上去错综复杂,简直了……
刷完题之后我很快地看了官方解,它的第一个思路就呼我一脸……这道题虽然看上去灯数很多,但是说到底就是6个灯的事儿,只要研究好一组6个灯的情况,就可以输出整体的情况。判断的方法应该跟我原来的思路差不太多,但是代码上清晰很多。
以下就是我的代码:
#include <iostream>
#include <fstream>
using namespace std;
const int MAXN = 101;
int s24 = 0;
int s3 = 0;
int s2 = 0;
int s34 = 0;
int N, C;
int res[16][MAXN] = {0};
int resLen = 0;
bool omits[16] = {0};
bool initData()
{
ifstream fin ("lamps.in");
fin >> N >> C;
int t;
for (fin >> t; t != -1; fin >> t) {
if (t % 6 == 1)
s24 = 2;
else if (t % 6 == 4)
s34 = 2;
else if (t % 2 == 0)
s3 = 2;
else
s2 = 2;
}
for (fin >> t; t != -1; fin >> t) {
if (t % 6 == 1) {
if (s24 != 2)
s24 = 1;
else
return false;
}
else if (t % 6 == 4) {
if (s34 != 2)
s34 = 1;
else
return false;
}
else if (t % 2 == 0) {
if (s3 != 2)
s3 = 1;
else
return false;
}
else {
if (s2 != 2)
s2 = 1;
else
return false;
}
}
return true;
}
void findCombos()
{
int i, j, k, m, n;
for (i = 0; i < 2; i ++) {
for (j = 0; j < 2; j ++) {
for (k = 0; k < 2; k ++) {
for (m = 0; m < 2; m ++) {
if (i+j+k+m > C)
continue;
if ((s24 && (s24+i+j+m)%2)
|| (s3 && (s3+i+k)%2)
|| (s2 && (s2+i+j)%2)
|| (s34 && (s34+i+k+m)%2))
continue;
for (n = 1; n <= N; n ++)
res[resLen]
= (i + (n%2?j:k) + (n%3==1?m:0) + 1) % 2;
resLen ++;
}
}
}
}
}
void sortCombos()
{
int i, j, k, t;
bool isS;
for (i = 0; i < resLen - 1; i ++) {
for (j = 0; j < resLen - i - 1; j ++) {
isS = false;
for (k = 1; k <= N; k ++) {
if (res[j][k] < res[j+1][k])
break;
else if (res[j][k] > res[j+1][k]) {
isS = true;
break;
}
}
if (isS) {
for (k = 1; k <= N; k ++) {
t = res[j][k];
res[j][k] = res[j+1][k];
res[j+1][k] = t;
}
}
}
}
for (i = 0; i < resLen - 1; i ++) {
isS = true;
for (j = 1; j <= N; j ++) {
if (res[i][j] != res[i+1][j]) {
isS = false;
break;
}
}
if (isS)
omits[i] = 1;
}
}
int main()
{
ofstream fout ("lamps.out");
bool f;
f = initData();
if (!f) {
fout << "IMPOSSIBLE" << endl;
return 0;
}
findCombos();
sortCombos();
if (resLen == 0) {
fout << "IMPOSSIBLE" << endl;
return 0;
}
int i, j, k;
for (i = 0; i < resLen; i ++) {
if (!omits[i]) {
for (j = 1; j <= N; j ++)
fout << res[i][j];
fout << endl;
}
}
}
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