poj2312 Battle City 【暴力 或 优先队列+BFS 或 BFS】

题意：M行N列的矩阵。Y：起点，T：终点。S、R不能走，走B花费2，走E花费1.求Y到T的最短时间。

三种解法。♪(^∇^*)

//解法一：暴力

//157MS
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<cmath>
#include<queue>
#include<limits.h>
#define CLR(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
const int N = 305;
int a[N][N], b[N][N];
char s[N][N];
int m, n;

int main() {
    int i, j;
    while(~scanf("%d%d", &m, &n), n || m) {
        getchar();
        int sx, sy, ex, ey;
        for(i = 0; i < m; ++i) scanf("%s", s[i]);
        for(i = 0; i < m; ++i) {
            for(j = 0; j < n; ++j) {
                if(s[i][j] == 'R' || s[i][j] == 'S')
                    a[i][j] = inf;
                else if(s[i][j] == 'B')
                    a[i][j] = 2;
                else  a[i][j] = 1;
                if(s[i][j] == 'Y') {sx = i; sy = j;}
                else if(s[i][j] == 'T') {ex = i; ey = j;}
            }
        }
        CLR(b, inf);
        b[sx][sy] = 0;
        while(1) {
            bool f = 0;
            for(i = 0; i < m; ++i) {
                for(j = 0; j < n; ++j) {
                    if(b[i][j] == inf) continue;
                    if(i > 0 && b[i-1][j] > b[i][j] + a[i-1][j]) {
                        b[i-1][j] = b[i][j] + a[i-1][j];
                        f = 1;
                    }
                    if(i < m-1 && b[i+1][j] > b[i][j] + a[i+1][j]) {
                        b[i+1][j] = b[i][j] + a[i+1][j];
                        f = 1;
                    }
                    if(j > 0 && b[i][j-1] > b[i][j] + a[i][j-1]) {
                        b[i][j-1] = b[i][j] + a[i][j-1];
                        f = 1;
                    }
                    if(j < n-1 && b[i][j+1] > b[i][j] + a[i][j+1]) {
                        b[i][j+1] = b[i][j] + a[i][j+1];
                        f = 1;
                    }
                }
            }
            if(!f) break;
        }
        if(b[ex][ey] != inf) printf("%d
", b[ex][ey]);
        else printf("-1
");
    }
    return 0;
}

//解法二：bfs+优先队列

//16MS
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<cmath>
#include<queue>
#include<limits.h>
#define CLR(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
const int N = 305;
struct node {
    int x, y;
    int w;
    node(int x = 0, int y = 0, int w = 0):x(x),y(y),w(w){}
    bool operator < (const node&r) const {
        return w > r.w;
    }
};
char s[N][N];
bool vis[N][N];
int dx[] = {1,0,-1,0};
int dy[] = {0,1,0,-1};
int m, n;
int ans;
int sx, sy;
bool check(int x, int y) {
    if(x < 0 || x >= m || y < 0 || y >= n)
        return false;
    if(s[x][y] == 'S' || s[x][y] == 'R' || vis[x][y])
        return false;
    return true;
}
int bfs() {
    CLR(vis, 0);
    priority_queue<node> q;
    q.push(node(sx, sy, 0));
    vis[sx][sy] = 1;
    while(!q.empty()) {
        node t = q.top(); q.pop();
        //vis[t.x][t.y] = 0;
        if(s[t.x][t.y] == 'T') {
            ans = t.w;
            return true;
        }
        for(int i = 0; i < 4; ++i) {
            int x = t.x + dx[i];
            int y = t.y + dy[i];
            if(check(x, y)) {
                int w = t.w + 1;
                if(s[x][y] == 'B') w++;
                vis[x][y] = 1;
                q.push(node(x, y, w));
            }
        }
    }
    return false;
}
int main() {
    int i, j;
    while(~scanf("%d%d", &m, &n), n || m) {
        getchar();
        for(i = 0; i < m; ++i) scanf("%s", s[i]);
        for(i = 0; i < m; ++i) {
            for(j = 0; j < n; ++j) {
                if(s[i][j] == 'Y') {sx = i; sy = j; break;}
            }
        }
        if(bfs()) printf("%d
", ans);
        else printf("-1
");
    }
    return 0;
}

//解法三：bfs，不用优先队列，搜到B时，把它变换成E再放入队列，重新取下一个队首继续搜索，这样就不用优先队列，用普通队列也能做啦

//32MS
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<cmath>
#include<queue>
#include<limits.h>
#define CLR(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
const int N = 305;
struct node {
    int x, y;
    int w;
    node(int x = 0, int y = 0, int w = 0):x(x),y(y),w(w){}
};
char s[N][N];
bool vis[N][N];
int dx[] = {1,0,-1,0};
int dy[] = {0,1,0,-1};
int m, n;
int ans;
int sx, sy;
bool check(int x, int y) {
    if(x < 0 || x >= m || y < 0 || y >= n)
        return false;
    if(s[x][y] == 'S' || s[x][y] == 'R' || vis[x][y])
        return false;
    return true;
}
int bfs() {
    CLR(vis, 0);
    queue<node> q;
    q.push(node(sx, sy, 0));
    vis[sx][sy] = 1;
    while(!q.empty()) {
        node t = q.front(); q.pop();
        if(s[t.x][t.y] == 'T') {
            ans = t.w;
            return true;
        }
        if(s[t.x][t.y] == 'B') {
            t.w++;
            s[t.x][t.y] = 'E';
            q.push(t);
            continue;
        }
        for(int i = 0; i < 4; ++i) {
            int x = t.x + dx[i];
            int y = t.y + dy[i];
            if(check(x, y)) {
                int w = t.w + 1;
                vis[x][y] = 1;
                q.push(node(x, y, w));
            }
        }
    }
    return false;
}
int main() {
    int i, j;
    while(~scanf("%d%d", &m, &n), n || m) {
        getchar();
        for(i = 0; i < m; ++i) scanf("%s", s[i]);
        for(i = 0; i < m; ++i) {
            for(j = 0; j < n; ++j) {
                if(s[i][j] == 'Y') {sx = i; sy = j; break;}
            }
        }
        if(bfs()) printf("%d
", ans);
        else printf("-1
");
    }
    return 0;
}