UVA 10047 The Monocycle

以前漏下的题目，现在重新做发现很水。
题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=105&page=show_problem&problem=988
独轮车问题，注意的就是vis数组的使用，裸的BFS就可以。vis数组是对当前坐标，颜色，方向的记录。为了熟悉一下优先队列，程序中使用了优先队列代替普通的队列，程序中可以在获得下一个状态时加入旋转180°的，此时step+=2就可以了。

#include <algorithm>#include <iostream>#include <string.h>#include <cstdlib>#include <iomanip>#include <sstream>#include <fstream>#include <climits>#include <cstdio>#include <vector>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>using namespace std;const int INF = 0x7fffffff;const int MAXL = 30;const int MAXC = 5;const int MAXD = 4;typedef struct _Node{    int x,y,color,step,dir;}Node;struct Cmp{    bool operator ()(const Node a,const Node b)    {        return a.step > b.step;    }};char maze[MAXL][MAXL];bool vis[MAXL][MAXL][MAXC][MAXD];int dc[] = {0,1,2,3,4};//0--Greenint dd[] = {0,1,2,3};//0--N,1--E,2--S,3--Wint dx[] = {-1,0,1,0};int dy[] = {0,1,0,-1};int L,R,Sx,Sy,Tx,Ty;priority_queue < Node ,vector <Node>, Cmp> Q;void Print_Node(Node cur){    printf("x = %d, y = %d, color = %d,step = %d,dir = %d\n",cur.x,cur.y,cur.color,cur.step,cur.dir);}bool IsOver(Node cur){    return cur.x == Tx && cur.y == Ty && cur.color == 0;}bool IsInMaze(Node cur){    return cur.x >= 0 && cur.x < L && cur.y >= 0 && cur.y < R;}bool IsBlock(Node cur){    return maze[cur.x][cur.y] != '#';}bool IsVisit(Node cur){    return vis[cur.x][cur.y][cur.color][cur.dir];}Node GetNextNode(Node cur){    Node newnode;    newnode.x = cur.x + dx[ cur.dir ];    newnode.y = cur.y + dy[ cur.dir ];    newnode.color = (cur.color + 1) % MAXC;    newnode.dir = cur.dir;    newnode.step = cur.step + 1;    return newnode;}Node TurnLeft90(Node cur){    Node newnode;    newnode.x = cur.x;    newnode.y = cur.y;    newnode.color = cur.color;    newnode.dir = (((cur.dir)+MAXD) - 1) % MAXD;    newnode.step = cur.step + 1;    return newnode;}Node TurnRight90(Node cur){    Node newnode;    newnode.x = cur.x;    newnode.y = cur.y;    newnode.color = cur.color;    newnode.dir = (cur.dir + 1) % MAXD;    newnode.step = cur.step + 1;    return newnode;}int BFS(){    while(!Q.empty())   Q.pop();    memset(vis,false,sizeof(vis));    Node cur,newnode;    cur.x = Sx,cur.y = Sy,cur.color = dc[0],cur.step = 0,cur.dir = dd[0];    Q.push(cur);    vis[cur.x][cur.y][cur.color][cur.dir] = true;    while(!Q.empty())    {        cur = Q.top();        Q.pop();        if(IsOver(cur)) return cur.step;        newnode = GetNextNode(cur);        if(IsInMaze(newnode) && IsBlock(newnode) && !IsVisit(newnode))        {            Q.push(newnode);            vis[newnode.x][newnode.y][newnode.color][newnode.dir] = true;        }        newnode = TurnLeft90(cur);        if(IsInMaze(newnode) && IsBlock(newnode) && !IsVisit(newnode))        {            Q.push(newnode);            vis[newnode.x][newnode.y][newnode.color][newnode.dir] = true;        }        newnode = TurnRight90(cur);        if(IsInMaze(newnode) && IsBlock(newnode) && !IsVisit(newnode))        {            Q.push(newnode);            vis[newnode.x][newnode.y][newnode.color][newnode.dir] = true;        }    }    return -1;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);#endif    int cases = 0;    while(~scanf("%d %d",&L,&R) && L+R)    {        gets(maze[0]);        for(int i = 0;i < L;i++)        {            gets(maze[i]);            for(int j = 0;j < R;j++)            {                if(maze[i][j] == 'S')                    Sx = i,Sy = j;                else if(maze[i][j] == 'T')                    Tx = i,Ty = j;            }        }        if(cases)   puts("");        printf("Case #%d\n",++cases);        int ans = BFS();        if(ans == -1)            printf("destination not reachable\n");        else            printf("minimum time = %d sec\n",ans);    }    return 0;}

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