Hdu 2579 Dating with girls(2) && hdu 2653 Waiting ten thousand years for Love【Bfs】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2653

                    http://acm.hdu.edu.cn/showproblem.php?pid=2579

题目的意思都很简单，主要写这篇博客的原因是学会了状态标记这个技巧，对于迷宫各种类 的，每个位置可以多次走过 的且对于每个格子的状态较少，可以多加一维来进行状态标记。

2653，对于每个位置，其余数就是一个状态。。比较简单。。一开始，自己想的各种，都是错的。。后来有了这个个技巧，马上就AC了。。这绝对的一个很好的技巧

Code：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>using namespace std;const int N = 85;const int dx[] = {0, 0, 1, -1};const int dy[] = {1, -1, 0, 0};int n, m, t, p, bx, by;char map[N][N];bool vis[N][N][N];//mark for the states...struct Node{    int x, y;    int ener, step;    Node(){ }    Node(int a, int b, int c, int d){        x = a; y = b;        ener = c; step = d;    }    bool friend operator < (Node X, Node Y){        if(X.step > Y.step) return true;        return false;    }};void bfs(){    memset(vis, 0, sizeof(vis));    priority_queue<Node> q;    q.push(Node(bx, by, p, 0));    vis[bx][by][p] = true;    while(!q.empty()){        Node tmp1 = q.top(), tmp2;        if(tmp1.step > t) break;        q.pop();        if(map[tmp1.x][tmp1.y] == 'L'){            if(tmp1.step <= t) printf("Yes, Yifenfei will kill Lemon at %d sec.\n", tmp1.step);            else puts("Poor Yifenfei, he has to wait another ten thousand years.");            return ;        }//        printf("%d %d %d %d\n", tmp1.x, tmp1.y, tmp1.step, tmp1.ener);        for(int i = 0; i < 4; i ++){            tmp2.x = tmp1.x + dx[i]; tmp2.y = tmp1.y + dy[i];            if(tmp2.x < 1 || tmp2.x > n || tmp2.y < 1 || tmp2.y > m || map[tmp2.x][tmp2.y] == '#') continue;            if(tmp1.ener != 0 && !vis[tmp2.x][tmp2.y][tmp1.ener - 1]){ // fly                tmp2.ener = tmp1.ener - 1;                tmp2.step = tmp1.step + 1;                vis[tmp2.x][tmp2.y][tmp2.ener] = true;                q.push(tmp2);            }// walk..            if(map[tmp1.x][tmp1.y] != '@' && map[tmp2.x][tmp2.y] != '@' && !vis[tmp2.x][tmp2.y][tmp1.ener]){                tmp2.ener = tmp1.ener;                tmp2.step = tmp1.step + 2;                vis[tmp2.x][tmp2.y][tmp2.ener] = true;                q.push(tmp2);            }        }    }    puts("Poor Yifenfei, he has to wait another ten thousand years.");    return ;}int main(){// you can step on the same gird if you have the different power    int k = 0;//    freopen("11.txt", "r", stdin);    while(~scanf("%d %d %d %d", &n, &m, &t, &p)){        getchar();        for(int i = 1; i <= n; i ++){            for(int j = 1; j <= m; j ++){                scanf("%c", &map[i][j]);                if(map[i][j] == 'Y') bx = i, by = j;            }            getchar();        }        printf("Case %d:\n", ++ k);        bfs();    }    return 0;}

另外一道题，也是很easy的题目，同样，状态很少。。

Code：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>using namespace std;const int N = 105;const int M = 15;const int dx[] = {0, 0, 1, -1};const int dy[] = {1, -1, 0, 0};char map[N][N];bool vis[N][N][M];int n, m, k, bx, by;struct Node{    int x, y;    int step;    Node(){ }    Node(int a, int b, int c){        x = a;        y = b;        step = c;    }    bool friend operator < (Node X, Node Y){        if(X.step > Y.step) return true;        return false;    }};void bfs(){    memset(vis, 0, sizeof(vis));    priority_queue<Node> q;    q.push(Node(bx, by, 0));    vis[bx][by][0] = true;    while(!q.empty()){        Node tmp1 = q.top(), tmp2;        q.pop();        if(map[tmp1.x][tmp1.y] == 'G'){            printf("%d\n", tmp1.step);            return ;        }//        printf("%d %d %d\n", tmp1.x, tmp1.y, tmp1.step);        for(int i = 0; i < 4; i ++){            tmp2.x = tmp1.x + dx[i]; tmp2.y = tmp1.y + dy[i];            if(tmp2.x < 1 || tmp2.x > n || tmp2.y < 1 || tmp2.y > m) continue;            if(map[tmp2.x][tmp2.y] == '#' && (tmp1.step + 1) % k == 0){                tmp2.step = tmp1.step + 1;                q.push(tmp2);            }            else if(map[tmp2.x][tmp2.y] != '#' && !vis[tmp2.x][tmp2.y][(tmp1.step + 1) % k])            {                    tmp2.step = tmp1.step + 1;                    vis[tmp2.x][tmp2.y][tmp2.step % k] = true;                    q.push(tmp2);            }        }    }    puts("Please give me another chance!");    return ;}int main(){//    freopen("1.txt", "r", stdin);    int T;    scanf("%d", &T);    while(T --){        scanf("%d %d %d", &n, &m, &k);        getchar();        for(int i = 1; i <= n; i ++){            for(int j = 1; j <= m; j ++){                scanf("%c", &map[i][j]);                if(map[i][j] == 'Y') bx = i, by = j;            }            getchar();        }        bfs();    }    return 0;}

应该尽快吧这几十道题目A了。。留给我们的时间确实已经不多了。。