poj 1167 dfs

为什么启发式搜索的题书上说是迭代加深？？？

思路跟大多数题一样，都是先把原问题进行处理，把处理过的数据再抽象成算法问题，因为原问题不好直接深搜，所以要先处理

注意启发式剪枝的>=ans就剪枝，而不是>才剪枝
注意题目上说了可以有多条线路的时间是一样的（两条线路起点都是3，间隔都是5）

#include <cstdio>
#include <iostream>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define IMAX 2147483646
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int

using namespace std;

#include <cstdio>
#include <iostream>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define IMAX 2147483646
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int

using namespace std;

struct Node {
	int beg, l, times;
	bool operator<(const Node &b)const {
		return times > b.times;
	}
}a[3333];
int n, t, cnt[333],tot,ans;

int pd(int beg, int l) {
	int ts = 0;
	while (beg < 60) {
		if (!cnt[beg])return 0;
		beg += l;
		ts++;
	}
	return ts;
}

void dfs(int x, int now, int sum) {
	if (now == n) {
		ans = min(ans, sum);
		return;
	}
	for (int i = x; i < tot; i++) {
		if (sum + (n - now) / a[i].times >= ans)break;//启发式剪枝
		if (!pd(a[i].beg, a[i].l))continue;//正确性判断
		for (int j = a[i].beg; j < 60; j += a[i].l)
			cnt[j]--;
		dfs(i, now + a[i].times, sum + 1);
		for (int j = a[i].beg; j < 60; j += a[i].l)
			cnt[j]++;
	}
}

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%d", &t), cnt[t]++;
	for(int i = 0;i<60;i++)
		for (int j = i + 1; j < 60 - i; j++) 
			if (pd(i, j))a[tot].beg = i, a[tot].l = j, a[tot++].times = pd(i, j);
	sort(a, a + tot);
	ans = 17;
	dfs(0, 0, 0);
	printf("%d\n", ans);
	return 0;
}