迭代加深搜索基础
题目描述
A little known fact about Bessie and friends is that they love stair climbing races. A better known fact is that cows really don’t like going down stairs. So after the cows finish racing to the top of their favorite skyscraper, they had a problem. Refusing to climb back down using the stairs, the cows are forced to use the elevator in order to get back to the ground floor.
The elevator has a maximum weight capacity of W (1 <= W <= 100,000,000) pounds and cow i weighs C_i (1 <= C_i <= W) pounds. Please help Bessie figure out how to get all the N (1 <= N <= 18) of the cows to the ground floor using the least number of elevator rides. The sum of the weights of the cows on each elevator ride must be no larger than W.
给出n个物品，体积为w[i]，现把其分成若干组，要求每组总体积<=W，问最小分组。(n<=18)
输入输出格式
输入格式：
-
Line 1: N and W separated by a space.
-
Lines 2..1+N: Line i+1 contains the integer C_i, giving the weight of one of the cows.
输出格式：
-
Line 1: A single integer, R, indicating the minimum number of elevator rides needed.
-
Lines 2..1+R: Each line describes the set of cows taking
one of the R trips down the elevator. Each line starts with an integer giving the number of cows in the set, followed by the indices of the individual cows in the set.
输入输出样例
输入样例#1： 复制
4 10
5
6
3
7
输出样例#1： 复制
3
2 1 3
1 2
1 4
说明
There are four cows weighing 5, 6, 3, and 7 pounds. The elevator has a maximum weight capacity of 10 pounds.
We can put the cow weighing 3 on the same elevator as any other cow but the other three cows are too heavy to be combined. For the solution above, elevator ride 1 involves cow #1 and #3, elevator ride 2 involves cow #2, and elevator ride 3 involves cow #4. Several other solutions are possible for this input.
思路
- 迭代加深搜索：这是一种类似广搜的深搜，但是它不需要广搜如此大的空间，它的空间与深搜的空间一样
可以看做带深度限制的DFS。
首先设置一个搜索深度，然后进行DFS，当目前深度达到限制深度后验证当前方案的合理性，更新答案。
不断调整搜索深度，直到找到最优解。
本题中枚举电梯数num,就是搜索的深度 - 剪枝: 可证第i奶牛放到i后车厢没有意义
代码
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define re register int
using namespace std;
int n, m, c[19], tot(0), ans(0), v[19];
bool dfs(int x, int num){
for(re i=1;i<=x&&i<=num;++i)
if(v[i]+c[x]<=m){
v[i]+=c[x];
if(x==n) return 1;
if(dfs(x+1,num)) return 1;
v[i]-=c[x];
}
return 0;
}
int main(){
scanf("%d%d",&n,&m);
for(re i=1;i<=n;++i) scanf("%d", &c[i]);
for(re i=1;i<=n;++i){//枚举厢数
memset(v,0,sizeof(v));
if (dfs(1,i)){
printf("%d
",i);
break;
}
}
return 0;
}