hdu1026 Ignatius and the Princess I（bfs+路径）

http://acm.hdu.edu.cn/showproblem.php?pid=1026

题意：求最短路长度、路径，有打怪操作，同样在路径中输出。

思路：很好的练习题吧，今天(2016/10/24)第二次做，打怪地方的标记初始化放在了结构体中，半天找不出错。好在改好了，1A~。

#include <stdio.h>#include <algorithm>#include <stdlib.h>#include <string.h>#include <iostream>#include <queue>#include <stack>#include <ctype.h>using namespace std;typedef long long LL;const int N = 105;const int INF = 0x3f3f3f3f;char G[N][N];bool vis[N][N];int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};int n, m;struct node{    int x, y, step, prex, prey;    int flag;    friend bool operator < (const node &a, const node &b)    {        return a.step > b.step;    }};node path[N][N];bool check(int x, int y){    if(x>=0 && x<n && y>=0 && y<m && !vis[x][y] && G[x][y]!='X') return true;    else return false;}void bfs(int x, int y){    memset(vis, false, sizeof(vis));    priority_queue<node>que;    node s;    s.x = x;    s.y = y;    s.step = 0;    vis[x][y] = true;    que.push(s);    while(!que.empty())    {        node tmp = que.top();        que.pop();        if(tmp.x==n-1 && tmp.y==m-1)        {            path[tmp.x][tmp.y] = tmp;            break;        }        for(int i = 0; i < 4; i++)        {            node tmp2;            tmp2 = tmp;            tmp2.x += dir[i][0];            tmp2.y += dir[i][1];            tmp2.step += 1;            if(check(tmp2.x, tmp2.y))            {                vis[tmp2.x][tmp2.y] = true;                if(isdigit(G[tmp2.x][tmp2.y]))                {                    tmp2.step += (G[tmp2.x][tmp2.y]-'0');                    tmp2.flag = (G[tmp2.x][tmp2.y]-'0');                }                else tmp2.flag = 0;//注意这里对flag赋值，不能在初始化结构体中赋值，不然输出有错                tmp2.prex = tmp.x;                tmp2.prey = tmp.y;                que.push(tmp2);                path[tmp2.x][tmp2.y] = tmp2;            }        }    }}void Print(){    stack<node>sta;    node now = path[n-1][m-1];    if(now.step == 0) printf("God please help our poor hero.\n");    else    {        sta.push(now);        printf("It takes %d seconds to reach the target position, let me show you the way.\n", now.step);        while(1)        {            now = path[now.prex][now.prey];            sta.push(now);            if(now.x==0 && now.y==0) break;        }        int time = 1;        sta.pop();        while(!sta.empty())        {            now = sta.top();            sta.pop();            if(now.flag == 0)                printf("%ds:(%d,%d)->(%d,%d)\n", time++, now.prex, now.prey, now.x, now.y);            else            {                printf("%ds:(%d,%d)->(%d,%d)\n", time++, now.prex, now.prey, now.x, now.y);                for(int i = 1; i <= now.flag; i++)                {                    printf("%ds:FIGHT AT (%d,%d)\n", time++, now.x, now.y);                }            }        }    }    printf("FINISH\n");}int main(){   // freopen("in.txt", "r", stdin);    while(~scanf("%d%d", &n, &m))    {        memset(path, 0, sizeof(path));        for(int i = 0; i < n; i++)            for(int j = 0; j < m; j++)            {                cin >> G[i][j];            }        bfs(0, 0);        Print();    }    return 0;}