【Fzu】2150 Fire Game（BFS）

来源：互联网发布：艺考软件编辑：程序博客网时间：2024/05/17 00:45

先判断连通性，然后选择两个点进行BFS，最终时间比一下大小即可，AC时间是1s刚刚好过了。

#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<queue>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;char mz[12][12];bool vis1[12][12];int vis[12][12];int n, m;int cnt;int Div[4][2] = { 0, 1, 0, -1, 1, 0, -1, 0 };bool H[12][12][12][12];struct pos{    int x, y;    int steps;    bool friend operator<(pos a, pos b)    {        return a.steps>b.steps;    }};bool check(int x, int y){    if (x >= 0 && x < n&&y >= 0 && y < m&&mz[x][y] == '#')        return true;    return false;}//判断连通性void dfs(int x, int y,int id){    for (int i = 0; i < 4; i++)    {        int xx = x + Div[i][0];        int yy = y + Div[i][1];        if (!vis[xx][yy] && check(xx, yy))        {            vis[xx][yy] = id;            dfs(xx, yy,id);        }    }}int bfs(int x, int y,int x1,int y1){    memset(vis1, 0, sizeof(vis1));    priority_queue<pos>q;    pos p, mid;    p.x = x;    p.y = y;    p.steps = 0;    q.push(p);    p.x = x1;    p.y = y1;    q.push(p);    int ans = 0;    vis1[x][y] = 1;    vis1[x1][y1] = 1;    while (!q.empty())    {        p = q.top();        q.pop();        if (p.steps > ans)            ans = p.steps;        for (int i = 0; i < 4; i++)        {            int xx = p.x + Div[i][0];            int yy = p.y + Div[i][1];            if (!vis1[xx][yy] && check(xx, yy))            {                vis1[xx][yy] = 1;                mid.x = xx;                mid.y = yy;                mid.steps = p.steps + 1;                q.push(mid);            }        }    }    return ans;}int main(){    int t;    cin >> t;    int icase = 1;    while (t--)    {        cin >> n >> m;        cin.get();        for (int i = 0; i < n; i++)            scanf("%s", mz[i]);        int id = 0;        memset(vis, 0, sizeof(vis));        memset(H, 0, sizeof(H));        for (int i = 0; i < n;i++)        for (int j = 0; j < m;j++)        if (mz[i][j] == '#'&&!vis[i][j])        {            vis[i][j] = ++id;            dfs(i, j, id);        }        int ans = 0x7ffffff;        for (int i = 0; i<n;i++)        for (int j = 0; j < m; j++)        {            if (mz[i][j] == '#')            for (int k = i; k < n; k++)            for (int l = 0; l<m; l++)            if ((vis[i][j] != vis[k][l] || id == 1) && mz[k][l] == '#'&&!H[i][j][k][l])            {                H[i][j][k][l] = H[k][l][i][j] = 1;                ans = min(ans, bfs(i, j, k, l));            }        }        cout << "Case " << icase++ << ": ";        if (id>2)            cout << "-1" << endl;        else            cout << ans << endl;    }}

较快的代码，留着思考

#include <string>#include <queue>#include <cstring>using namespace std;int n, m, ret, res;string mz[11];int vis[11][11];queue<pair<int, int > > q;int dir[4][2] = {-1,0,1,0,0,1,0,-1};void bfs(int x1, int y1, int x2, int y2) {    memset(vis, 0, sizeof(vis));    while(!q.empty()) q.pop();    q.push(make_pair(x1,y1));    q.push(make_pair(x2,y2));    vis[x1][y1] = 1;    vis[x2][y2] = 1;    ret = 1;    while(!q.empty()) {        pair<int, int> now = q.front(); q.pop();        int x = now.first;        int y = now.second;        for (int i = 0; i < 4; ++i) {            int tx = x+dir[i][0];            int ty = y+dir[i][1];            if (tx < 0 || ty < 0 || tx >= n || ty >= m) continue;            if (mz[tx][ty] != '#') continue;            if (vis[tx][ty]) continue;            vis[tx][ty] = vis[x][y] + 1;            if (ret < vis[tx][ty]) ret = vis[tx][ty];            q.push(make_pair(tx,ty));        }    }    return ;}void dfs(int x, int y, int ct) {    for (int i = 0; i < 4; ++i) {        int tx = x+dir[i][0];        int ty = y+dir[i][1];        if (tx < 0 || ty < 0 || tx >= n || ty >= m) continue;        if (mz[tx][ty] != '#') continue;        if (vis[tx][ty]) continue;        vis[tx][ty] = ct;        dfs(tx,ty,ct);    }    return ;}bool ok() {    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            if (mz[i][j] == '#' && !vis[i][j])                 return false;        }    }    return true;}void output() {    for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j)             cout << vis[i][j] << " ";        cout << endl;    }    return ;}bool mmvs[11][11][11][11];int main() {    int t;    cin >> t;    for (int kase = 1; kase <= t; ++ kase) {        cin >> n >> m;        for (int i = 0; i < n; ++i) {            cin >> mz[i];        }        int ct = 0;        memset(vis,0,sizeof(vis));        memset(mmvs,0,sizeof(mmvs));        for (int i = 0; i < n; ++i) {            for (int j = 0; j < m; ++j) {                if (!vis[i][j] && mz[i][j] == '#') {                    vis[i][j] = ++ct;                    dfs(i, j, ct);                }            }        }        res = 0x7fffffff;        cout << "Case " << kase << ": ";        if (ct > 2) { cout << "-1" << endl; continue;}        for (int i = 0; i < n; ++i) {            for (int j = 0; j < m; ++j) {                if (mz[i][j] != '#') continue;                for (int p = 0; p < n; ++p) {                    for (int q = 0; q < m; ++q) {                        if (mz[p][q] != '#') continue;                        //if (i==p && j ==q) continue;                        if(mmvs[i][j][p][q]) continue;                        mmvs[i][j][p][q] = mmvs[p][q][i][j] = true;                        bfs(i,j,p,q);                        if (ok() && ret-1 < res) res = ret-1;                        //output();                    }                }            }        }        cout << res << endl;    }    return 0;}

0 0