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1.hdoj 1728 逃离迷宫 

这题本来用优先队列做的，WA很多次。 网上看到下面的解释： 

举个例子，到达某个点的时候方向为向上或者向右的最少转弯次数都是4次
假如终点在向上的方向，那么总的最少转弯次数就是4次
但是开2维数组有可能造成记录的是向右的状态，因为同是4次。
那么这样到达终点就需要5次转弯了。

 
#include <iostream>
#include <queue>
#include <cstdio>
using namespace std;

class pt{
public:
    int x, y;
};

int dir[4][2] = {1, 0, 0, 1, 0, -1, -1, 0};
char mp[110][110];
int num[110][110];
int limit, sx, sy, ex, ey, n, m;

bool solve(){
    pt p, q;
    int i, t;
    queue<pt> Q;
    memset(num, -1, sizeof(num));
    p.x = sx;
    p.y = sy;
    Q.push(p);
    while(!Q.empty()){
        p = Q.front();
        Q.pop();
        t = num[p.x][p.y] + 1;
        if(t > limit) continue;
        for(i=0; i<4; ++i){
            q.x = p.x + dir[i][0];
            q.y = p.y + dir[i][1];
            while(q.x >= 0 && q.x < n && q.y >= 0 && q.y < m && mp[q.x][q.y]=='.'){
                if(num[q.x][q.y] == -1 || num[q.x][q.y] >= t){
                    num[q.x][q.y] = t;
                    Q.push(q);
                    if(q.x == ex && q.y == ey) return 1;
                }
                q.x += dir[i][0];
                q.y += dir[i][1];
            }
        }
    }
    return 0;
}

void init(){
    int i;
    scanf("%d %d", &n, &m);
    for(i=0; i<n; ++i){
        scanf("%s", mp[i]);
    }
    scanf("%d %d %d %d %d", &limit, &sy, &sx, &ey, &ex);
    --sx; --sy;
    --ex; --ey;
}

int main(){
//    freopen("c:/aaa.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--){
        init();
        if(solve()) puts("yes");
        else puts("no");
    }
    return 0;
}

2.hdoj 1242 Rescue

以Angel为出发点，找出“距离”最近的朋友。

 
#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#define inf 10000000
using namespace std;

class Node{
public:
    int x, y, v;
};

bool operator>(Node a, Node b){
    return a.v > b.v;
}

int n, m;
int sx, sy;
int dir[4][2] = {1, 0, 0, 1, 0, -1, -1, 0};
char mp[210][210];
int num[210][210];

void solve(){
    int i, j, ans;
    priority_queue<Node, vector<Node>, greater<Node> > Q;
    int mark[210][210];
    Node p, q;
    memset(mark, 0, sizeof(mark));
    memset(num, 0, sizeof(num));

    p.x = sx;
    p.y = sy;
    p.v = 0;
    Q.push(p);
    mark[sx][sy] = 1;

    while(!Q.empty()){
        p = Q.top();
        Q.pop();
        for(i=0; i<4; ++i){
            q.x = p.x + dir[i][0];
            q.y = p.y + dir[i][1];

            if(q.x < 0 || q.x >= n || q.y < 0 || q.y >= m) continue;
            if( mp[q.x][q.y] == '#' || mark[q.x][q.y] ) continue;

            q.v = p.v + 1;
            if(mp[q.x][q.y] == 'x') ++q.v;

            Q.push(q);
            mark[q.x][q.y] = 1;
            num[q.x][q.y] = q.v;
        }
    }

    ans = inf;
    for(i=0; i<n; ++i){
        for(j=0; j<m; ++j){
            if(mp[i][j] == 'r' && num[i][j] < ans && mark[i][j])
                ans = num[i][j];
        }
    }

    if(ans == inf) puts("Poor ANGEL has to stay in the prison all his life.");
    else printf("%d\n", ans);
}

void init(){
    int i, j;
    for(i=0; i<n; ++i){
        scanf("%s", mp[i]);
    }

    for(i=0; i<n; ++i){
        for(j=0; j<m; ++j){
            if(mp[i][j] == 'a'){
                sx = i;
                sy = j;
                break;
            }
        }
    }
}

int main(){
//    freopen("c:/aaa.txt", "r", stdin);
    while(scanf("%d %d", &n, &m)!=EOF){
        init();
        solve();
    }
    return 0;
}