SCU2016-04 B题 枚举+ bfs最短路

Analyse:
用bfs处理相互之间的最短路，然后dfs枚举这个路径的所有情况。
复杂度：O(n∗m∗k!)
Get:
数据小的，尽管暴力就是了。

/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 一  7/11 09:59:19 2016**Problem**:**Code**:***********************[e-mail]**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;typedef vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a))#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3f3f3f3f;const int maxn = 1e5 + 13;char G[109][109];int n, m, k;int sx, sy;int dirx[] = {-1, 0, 1, 0};int diry[] = {0, 1, 0, -1};struct node {    int x, y, step;    node(){}    node(int a, int b, int c): x(a), y(b), step(c){}};bool vis[109][109];int bfs(int sx, int sy, int ex, int ey){    memset(vis, false, sizeof(vis));    queue<node> que;    int ret = INF;    node t = node(sx, sy, 0);    que.push(t);    while (!que.empty()) {        node cur = que.front(); que.pop();        if (cur.x == ex && cur.y == ey) {            ret = cur.step;            break;        }        vis[cur.x][cur.y] = true;        for (int i = 0; i < 4; i++) {            node nxt(cur.x + dirx[i], cur.y + diry[i], cur.step + 1);            if (nxt.x > 0 && nxt.x <= n && nxt.y > 0 && nxt.y <= m && (G[nxt.x][nxt.y] == '.')) {                if (!vis[nxt.x][nxt.y]) que.push(nxt);                vis[nxt.x][nxt.y] = true;            }        }    }    return ret;}pii aim[5];int d[5][5];bool vv[10];int ans = INF;void dfs(int s, int l, int par) {    if (l >= ans) return;    if (s == k) {        ans = min(ans, l);        return;    }    for (int i = 0; i < k; i++) {        if (!vv[i]) {            vv[i] = true;            dfs(s + 1, l + d[par][i], i);            vv[i] = false;        }    }}int main(void){#ifdef LOCAL    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    while (scanf("%d%d", &n, &m)) {        ans = INF;        if (n == 0 && m == 0) break;        for (int i = 1; i <= n; i++) {            getchar();            for (int j = 1; j <= m; j++) {                scanf("%c", &G[i][j]);                if (G[i][j] == '@') G[i][j] = '.', sx = i, sy = j;            }        }        sa(k);        for (int i = 0; i < k; i++) {            scanf("%d%d", &aim[i].xx, &aim[i].yy);         }        aim[k].xx = sx, aim[k].yy = sy;        for (int i = 0; i <= k; i++) {            for (int j = 0; j <= k; j++) {                d[i][j] = bfs(aim[i].xx, aim[i].yy, aim[j].xx, aim[j].yy);            }        }        memset(vv, false, sizeof(vv));        dfs(0, 0, k);        pri(ans == INF ? -1 : ans);    }    return 0;}