hihoCoder 1233 Boxes（bfs）

来源：互联网发布：网络订花送花编辑：程序博客网时间：2024/05/31 05:27

题目链接：hihoCoder 1233 Boxes

解题思路

确定n，bfs预处理出所有可达状态。

代码

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;const int maxn = 10;const int maxs = 1e6;const int dir[2] = {1, -1};const int inf = 0x3f3f3f3f;typedef pair<int,int> pii;int N, E, V[maxn], P[maxn], T[maxn], vis[8][maxs];int idx(int* c) {    int ret = 0;    for (int i = 0; i < N; i++)        ret = ret * N + c[i];    return ret;}void ridx(int* c, int s) {    for (int i = N-1; i >= 0; i--) {        c[i] = s % N;        s /= N;    }}bool judge (int w, int u, int p) {    if (p < 0 || p >= N) return false;    return w < T[p] && w == T[u];}void bfs (int n, int* v) {    N = n;    for (int i = 0; i < N; i++) V[i] = T[i] = i;    int s = idx(T);    v[s] = 0;    queue<pii> Q;    Q.push(make_pair(s, 0));    while (!Q.empty()) {        int u = Q.front().first;        int c = Q.front().second;        Q.pop();        ridx(P, u);        for (int i = 0; i < N; i++) T[i] = inf;        for (int i = 0; i < N; i++) T[P[i]] = min(T[P[i]], V[i]);        for (int i = 0; i < N; i++) {            for (int j = 0; j < 2; j++) {                if (judge(V[i], P[i], P[i]+dir[j])) {                    P[i] += dir[j];                    int e = idx(P);                    if (v[e] == -1) {                        Q.push(make_pair(e, c+1));                        v[e] = c+1;                    }                    P[i] -= dir[j];                }            }        }    }}void init () {    scanf("%d", &N);    for (int i = 0; i < N; i++) scanf("%d", &V[i]);    memcpy(P, V, sizeof(V));    sort(P, P + N);    for (int i = 0; i < N; i++) T[i] = lower_bound(P, P + N, V[i]) - P;    for (int i = 0; i < N; i++) P[T[i]] = i;    //for (int i = 0; i < N; i++) printf("%d ", P[i]);    //printf("\n");    E = idx(P);}int main () {    int cas;    scanf("%d", &cas);    memset(vis, -1, sizeof(vis));    for (int i = 1; i < 8; i++) bfs(i, vis[i]);    while (cas--) {        init();        printf("%d\n", vis[N][E]);    }    return 0;}

0 0