uva 11624 Fire

注意：

（1）加上火之后，预处理出每个格子的最早着火时间，当Joe走到格子时，进行判断能否走即可。

（2）刚开始的着火点有多个，如何同时求在多个着火点的情况下的最早着火时间。

考虑建造一个超级源，连向出始得着火点，则一次广搜即可。

思考可以建造超级源的条件？？？

每个着火点的蔓延过程是相互独立的，且每个格子最小着火时间即所有着火点蔓延到该格子时间的最小值。

//#pragma warning (disable: 4786)
//#pragma comment (linker, "/STACK:16777216")
//HEAD
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FD(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)

typedef long long LL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-10;
const int maxn = 20;

typedef pair<int, int> pii;
queue<pii>Q;
pii st;
struct node{
    pii x;
    int dis;
    node(pii x, int dis): x(x), dis(dis){}
};
queue<node>q;
char s[1010];
int tab[1010][1010];
int n, m;
int dir_i[4] = {1, -1, 0, 0};
int dir_j[4] = {0, 0, 1, -1};
bool vis[1010][1010];
bool check(int x, int y)
{
    if (x < 0 || x >= n || y < 0 || y >= m) return false;
    return true;
}
int main ()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        while (!Q.empty()) Q.pop();
        while (!q.empty()) q.pop();
        memset(vis, 0, sizeof(vis));
        memset(tab, INF, sizeof(tab));
        scanf("%d%d", &n, &m);
        for (int i = 0; i < n; i++)
        {
            scanf("%s", s);
            for (int j = 0; j < m; j++)
            {
                if (s[j] == ‘#‘) tab[i][j] = 0;
                else if (s[j] == ‘F‘)
                {
                    Q.push(make_pair(i, j));
                    tab[i][j] = 0;
                }
                else if (s[j] == ‘J‘)
                    st = make_pair(i, j);
            }
        }

        while (!Q.empty())
        {
            pii x = Q.front(); Q.pop();
            for (int r = 0; r < 4; r++)
            {
                int ni = x.first + dir_i[r];
                int nj = x.second + dir_j[r];
                if (check(ni, nj) && tab[ni][nj] > tab[x.first][x.second] + 1)
                {
                    tab[ni][nj] = tab[x.first][x.second] + 1;
                    Q.push(make_pair(ni, nj));
                }
            }
        }

        q.push(node(st, 0));
        vis[st.first][st.second] = true;
        int fla = 0, ans;
        if (st.first == 0 || st.first == n - 1 || st.second == 0 || st.second == m - 1)
        {
            fla = 1;
            ans = 1;
        }

        while (!q.empty())
        {
            if (fla) break;
            node a = q.front(); q.pop();
            pii x = a.x;
            int nextx = a.dis + 1;
            for (int r = 0; r < 4; r++)
            {
                int ni = x.first + dir_i[r];
                int nj = x.second + dir_j[r];
                if (check(ni, nj) && !vis[ni][nj] && tab[ni][nj] > nextx)
                {
                    if (ni == 0 || ni == n - 1 || nj == 0 || nj == m - 1)
                    {
                        fla = 1;
                        ans = nextx + 1;
                        break;
                    }
                    vis[ni][nj] = true;
                    q.push(node(make_pair(ni, nj), nextx));
                }
            }
        }
        if (fla) printf("%d\n", ans);
        else puts("IMPOSSIBLE");
    }
    return 0;
}

uva 11624 Fire,布布扣,bubuko.com

uva 11624 Fire

原文：http://blog.csdn.net/guognib/article/details/21712581