HDU 2612 Find a way——bfs

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直接枚举@bfs会超时，正确做法是两遍bfs求出两个人到所有@的最短路，然后再遍历求解

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <queue>using namespace std;const int dx[] = {0, 0, -1, 1};const int dy[] = {-1, 1, 0, 0};const int INF = 0x3f3f3f3f;const int maxn = 200 + 10;char g[maxn][maxn];int n, m, cnt, vis[maxn][maxn];struct Dis {    int ok, dis;}dis[maxn][maxn];struct State {    int x, y, step;}state, s1, s2;void bfs(int x, int y) {    memset(vis, 0, sizeof(vis));    vis[x][y] = 1;    queue<State> q;    q.push(State{x, y, 0});    int num = 0;    while (!q.empty()) {        state = q.front(); q.pop();        if (g[state.x][state.y] == '@') {            num++;            if (dis[state.x][state.y].dis) dis[state.x][state.y].ok = 1;            dis[state.x][state.y].dis += state.step;        }        if (num == cnt) break;        for (int i = 0; i < 4; i++) {            int xx = state.x + dx[i], yy = state.y + dy[i];            if (1 <= xx && xx <= n && 1 <= yy && yy <= m && g[xx][yy] != '#' && !vis[xx][yy]) {                vis[xx][yy] = 1;                q.push(State{xx, yy, state.step + 1});            }        }    }}int main() {    while (scanf("%d %d", &n, &m) == 2) {        cnt = 0;        for (int i = 1; i <= n; i++) {            scanf("%s", g[i] + 1);            for (int j = 1; j <= m; j++) {                if (g[i][j] == 'Y') s1.x = i, s1.y = j;                else if (g[i][j] == 'M') s2.x = i, s2.y = j;                else if (g[i][j] == '@') cnt++;            }        }        memset(dis, 0, sizeof(dis));        bfs(s1.x, s1.y); bfs(s2.x, s2.y);        int ans = INF;        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= m; j++) {                if (dis[i][j].ok) {                    ans = min(ans, dis[i][j].dis);                }            }        }        printf("%d\n", ans * 11);    }    return 0;}

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