IDA*
这题真不会写。。估价函数太巧妙了。。

按照lyd神牛的说法我们把a[i+1]=a[i]+1记为正确后继，反之则记为错误后继

那么考虑最优的一次交换区间，至多能够纠正三个错误后继，所以我们统计序列的错误后继数n，n/3就是估价函数的值

因为把某区间移到后面和把另外一个区间移到它前面是等价的，所以我们按从左往右考虑区间后移即可

初始迭代深度为原序列最少需要的移动次数，移动次数+估价函数值超过迭代深度就直接返回搜索失败

每一次搜索完之后，最少移动次数（迭代深度）++

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}

int _, n, maxdepth;

int f(int a[]){
    int tot = 0;
    for(int i = 1; i <= n - 1; i ++) if(a[i + 1] != a[i] + 1) tot ++;
    if(a[n] != n) tot ++;
    return tot;
}
// swap [x...k] and [k+1...y]
void move(int a[], int x, int k, int y){
    int tmp[20], i = 0, j = 0;
    for(i = k + 1, j = 0; i <= y; i ++, j ++)
        tmp[j] = a[i];
    for(i = x; i <= k; i ++, j ++)
        tmp[j] = a[i];
    for(i = x, j = 0; i <= y; i ++, j ++)
        a[i] = tmp[j];
}

bool dfs(int a[], int step){
    int h = f(a);
    //cout << h << endl;
    if(h == 0) return true;
    if(h + 3 * step > maxdepth * 3) return false;
    for(int i = 1; i < n; i ++){
        for(int k = i; k < n; k ++){
            for(int j = k + 1; j <= n; j ++){
                int tmp[20];for(int pos = 1; pos <= n; pos ++) tmp[pos] = a[pos];
                //for(int x = 1; x <= n; x ++) printf("%d ", tmp[x]); puts("");
                move(tmp, i, k, j);
                //for(int x = 1; x <= n; x ++) printf("%d ", tmp[x]); puts("");
                if(dfs(tmp, step + 1)) return true;
            }
        }
    }
    return false;
}

void IDAstar(int a[]){
    maxdepth = (int)ceil((double)f(a) / 3);
    if(maxdepth) while(maxdepth < 5 && !dfs(a, 0)) maxdepth ++;
    if(maxdepth == 5) printf("5 or more\n");
    else printf("%d\n", maxdepth);
}

int main(){

    scanf("%d", &_);
    for(; _ > 0; _ --){
        scanf("%d", &n);
        int b[20]; for(int i = 1; i <= n; i ++) scanf("%d", &b[i]);
        IDAstar(b);
    }
    return 0;
}

标签：Booksort,return,int,lyd,ch,POJ,ans,inline,习题
来源： https://www.cnblogs.com/onionQAQ/p/10543330.html