題意及思路

模擬場上用一般方法枚舉非常麻煩，一個小時沒寫出來，還是自己太菜了。。。用表達式樹枚舉有一個好處，判斷需不需要加括號非常方便，只有當前節點運算符的優先級高於子節點的時候，才需要給子節點加一個括號。

代碼：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int cost[] = {0,1,1,2,2,3,1,2,2,3,3,4,2,3,3,4,4,5,3,4,4,5,5,6};
int ans = INF, num_cnt, flag_cnt;
int a[10], sum;
int b[5] = {0, 1, 2, 3, 4}, c[10];
struct node{
	int val, ls, rs;
};
node tr[110];
void init() {
 
　　 tr[1].ls=2;tr[1].rs=3;tr[2].ls=4;tr[2].rs=5;tr[3].ls=6;tr[3].rs=7;
    tr[8].ls=9;tr[8].rs=10;tr[9].ls=11;tr[9].rs=12;tr[11].ls=13;tr[11].rs=14;
    tr[15].ls=16;tr[15].rs=17;tr[16].ls=18;tr[16].rs=19;tr[19].ls=20;tr[19].rs=21;
    tr[22].ls=23;tr[22].rs=24;tr[24].ls=25;tr[24].rs=26;tr[25].ls=27;tr[25].rs=28;
    tr[29].ls=30;tr[29].rs=31;tr[31].ls=32;tr[31].rs=33;tr[33].ls=34;tr[33].rs=35;
}
bool is_leaf(int x) {
	return (tr[x].ls == 0) && (tr[x].rs == 0);
}
int op(int x, int y, int flag) {
	if(x == -INF || y == -INF) return -INF;
	if(flag == 0) return x + y;
	if(flag == 1) return x - y;
	if(flag == 2) return x * y;
	if((y == 0) || (flag == 3 && (x % y) != 0 )) return -INF;
	return x / y;
}
void build_tree(int x) {
	if(tr[x].ls == 0 && tr[x].rs == 0) {
		tr[x].val = a[b[++num_cnt]];
		return;
	} else {
		tr[x].val = c[++flag_cnt];
		build_tree(tr[x].ls);
		build_tree(tr[x].rs);
	}
}
int cal(int x) {
	if(tr[x].ls == 0 && tr[x].rs == 0) return tr[x].val;
	if(tr[x].val >= 2) {
		if(!is_leaf(tr[x].ls) && tr[tr[x].ls].val < 2) sum++;
		if(!is_leaf(tr[x].rs) && tr[tr[x].rs].val < 2) sum++;
	}
	return op(cal(tr[x].ls), cal(tr[x].rs), tr[x].val);
}
int main() {
	for (int i = 1; i <= 4; i++)
		scanf("%d", &a[i]);
	int now_state = 0;
	init();
	do {
		for (int t = 0; t < 5; t++) {
			for (int i = 0; i < 4; i++)
				for (int j = 0; j < 4; j++)
					for (int k = 0; k < 4; k++) {
						c[1] = i, c[2] = j, c[3] = k;
						flag_cnt = num_cnt = 0;
						sum = cost[now_state] * 2;
						build_tree(7 * t + 1);
						int now = cal(7 * t + 1);
						if(now == 24) {
							ans = min(ans, sum);
						}
					}
		}
		now_state++;
	}while(next_permutation(b + 1, b + 1 + 4));
	if(ans == INF) printf("impossible\n");
	else printf("%d\n", ans);
} 

gym - 101673I Twenty Four, Again (表達式樹枚舉)