POJ 2362 解题报告

之前程序有个bug（更新下次从下一个棒子找），一直没有发现，总是TLE，最后改着改着就和discuss里面这个post一样了：
http://poj.org/showmessage?message_id=177718
这道题是DFS+剪枝。能否通过，重点还是在于保证正确性前提下的剪枝。

将棒子按照长度从大到小排序。这样做的原因是：1.排序肯定不会成为瓶颈(O(nlogn) vs O(n!))。排序还可以排除出有一根特长的棒子的情况（超过了边长）。2.从大到小找可以fail faster.

1.如果找得只剩下一组了，我们就不用再找了。剩下的一组肯定能拼出来。

2.如果是找一组的第一个棒子失败了，那么就不必接着找了。因为当前尝试的这个棒子是肯定可以作为新的组的第一个棒子的，失败的原因只能是之前的棒子已经组合错了，需要回溯，而不是接着尝试下一个棒子。

3.如果在同一组中，当前尝试的棒子和之前的棒子长度相同，那么这个棒子也可以直接跳过了。因为和之前那个棒子没有区别，肯定也会失败。

thestoryofsnow

2362

Accepted

132K

16MS

C++

/*
ID: thestor1
LANG: C++
TASK: poj2362
*/
#include <iostream>
#include <fstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <limits>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <cassert>

using namespace std;

const int MAXM = 40;

int sticks[MAXM];
bool visited[MAXM];

int M, LEN;

const int MATCH = 4;

bool dfs(int len, int matched, int index)
{
int now = -1;
for (int i = index; i < M; ++i)
{
// if we have tried a stick with the same length and failed (we know from the fact that we are trying other sticks)
// we don't need to try the same length again
if (sticks[i] == now)
{
continue;
}

if (!visited[i] && sticks[i] <= len)
{
now = sticks[i];
visited[i] = true;
if (sticks[i] == len)
{
if (matched == MATCH - 2)
{
// since we already matched 2 times, plus this time,
// we are confident that whatever left is sufficient for the remaining last match
return true;
}
else if (dfs(LEN, matched + 1, 0))
{
return true;
}
else
{
// since can not find a new match from scratch,
// we don't need to try next stick
visited[i] = false;
return false;
}
}
else
{
if (dfs(len - sticks[i], matched, i + 1))
{
return true;
}
visited[i] = false;
if (len == LEN)
{
// since can not find a new match from scratch (and this stick should be part of a match),
// we don't need to try next stick
return false;
}
}
}
}
return false;
}

int main()
{
int N;
scanf("%d", &N);
for (int t = 0; t < N; ++t)
{
scanf("%d", &M);
int sum = 0;
for (int i = 0; i < M; ++i)
{
scanf("%d", &sticks[i]);
sum += sticks[i];
}

if (sum % 4 != 0 || M < 4)
{
printf("no\n");
}
else
{
LEN = sum / 4;
sort(sticks, sticks + M, greater<int>());

// printf("sticks:\n");
// for (int i = 0; i < M; ++i)
// {
// 	printf(" %d\n", sticks[i]);
// }
// printf("\n");

if (sticks[0] > LEN)
{
printf("no\n");
continue;
}

for (int i = 0; i < M; ++i)
{
visited[i] = false;
}

if (dfs(LEN, 0, 0))
{
printf("yes\n");
}
else
{
printf("no\n");
}

}
}
return 0;
}