HDU 1026 Ignatius and the Princess I （搜索）

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

很明显宽搜，有点像dijkstra算法，使用优先队列每次取一个最小的节点然后更新相邻节点。

到达终点直接退出，之后寻找路径-记录父节点或者从后往前找（应该都可以吧，我用的后者）

刚开始用的深搜直接超时，这几天写深搜写习惯了。脑残了。

代码：

#define _CRT_SECURE_NO_WARNINGS
#include <functional>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <cctype>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1000000007;
ll powmod(ll a, ll b) { ll res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res*a%mod; a = a*a%mod; }return res; }
// head
const int inf = 0x3f3f3f3f;
#define maxn 205
int iadd[] = {0, 1, 0, -1}, jadd[] = {1, 0, -1, 0};
struct rtnode{
    int i, j;
    int fight;
    rtnode(): i(0), j(0), fight(0) {}
    rtnode(int f, int t, int fi): i(f), j(t), fight(fi) {}
    bool operator < (const rtnode & t)const {
        return fight > t.fight;
    }
}bestrt[maxn * maxn];
int mv[maxn][maxn];
char maze[maxn][maxn];
int n, m;
bool flag;
priority_queue<rtnode> q;
stack<rtnode> s;

int bfs(){
    flag = false;
    memset(mv, -1, sizeof(mv));
    while(!s.empty())
        s.pop();
    while(!q.empty())
        q.pop();
    q.push(rtnode(0, 0, 0));
    mv[0][0] = 0;
    while(!q.empty()){
        rtnode tmnode = q.top();
        q.pop();
        
        if(tmnode.i == n - 1 && tmnode.j == m - 1){
            flag = true;
            break;
        }
        
        for(int i = 0; i < 4; i++){
            int nexti, nextj;
            nexti = tmnode.i + iadd[i]; nextj = tmnode.j + jadd[i];
            if(nexti < 0 || nextj < 0 || nexti >= n || nextj >= m)
                continue;
            if(mv[nexti][nextj] != -1 || maze[nexti][nextj] == 'X')
                continue;
            int tmpl = 1 + tmnode.fight;
            if(isdigit(maze[nexti][nextj])){
                tmpl += maze[nexti][nextj] - '0';
            }
            mv[nexti][nextj] = tmpl;
            q.push(rtnode(nexti, nextj, tmpl));
        }
    }
    if(!flag)
        return 0;
    s.push(rtnode(n - 1, m - 1, mv[n - 1][m - 1]));
    while(true){
        rtnode tmn = s.top();
        if(tmn.i == 0 && tmn.j == 0)
            break;
        for(int i = 0; i < 4; i++){
            int nexti, nextj;
            nexti = tmn.i + iadd[i]; nextj = tmn.j + jadd[i];
            if(nexti < 0 || nextj < 0 || nexti >= n || nextj >= m)
                continue;
            int tms = 1;
            if(isdigit(maze[tmn.i][tmn.j])){
                tms += maze[tmn.i][tmn.j] - '0';
            }
            if(mv[nexti][nextj] == tmn.fight - tms){
                s.push(rtnode(nexti, nextj, mv[nexti][nextj]));
                break;
            }
        }
    }
    for(int i = 0; !s.empty(); i++){
        bestrt[i] = s.top();
        bestrt[i].fight = 0;
        int tmi = bestrt[i].i, tmj = bestrt[i].j;
        if(isdigit(maze[tmi][tmj]))
            bestrt[i].fight = maze[tmi][tmj] - '0';
        s.pop();
    }
    return mv[n - 1][m - 1];
}

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                scanf(" %c", &maze[i][j]);
            }
        }
        
        int minstp = bfs();
        
        if(!flag){
            puts("God please help our poor hero.
FINISH");
        }
        else{
            printf("It takes %d seconds to reach the target position, let me show you the way.
", minstp);
            int cnt = 1;
            for(int i = 1; cnt <= minstp; i++){
                printf("%ds:(%d,%d)->(%d,%d)
", cnt, bestrt[i - 1].i, bestrt[i - 1].j, bestrt[i].i, bestrt[i].j);
                if(bestrt[i].fight != 0){
                    for(int k = 1; k <= bestrt[i].fight; k++){
                        printf("%ds:FIGHT AT (%d,%d)
", cnt + k, bestrt[i].i, bestrt[i].j);
                    }
                    cnt += bestrt[i].fight + 1;
                }
                else{
                    cnt++;
                }
            }
            printf("FINISH
");
            
        }
    }
}
/*
5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX.
5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX1
5 6
.XX...
..XX1.
2...X.
...XX.
XXXXX.
*/

题目：

Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18947    Accepted Submission(s): 6131
Special Judge

Problem Description

The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-top corner is (0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0), and the door to feng5166's room is at (N-1,M-1), that is our target. There are some monsters in the castle, if Ignatius meet them, he has to kill them. Here is some rules:

1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.

Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.

 

Input

The input contains several test cases. Each test case starts with a line contains two numbers N and M(2<=N<=100,2<=M<=100) which indicate the size of the labyrinth. Then a N*M two-dimensional array follows, which describe the whole labyrinth. The input is terminated by the end of file. More details in the Sample Input.

 

Output

For each test case, you should output "God please help our poor hero." if Ignatius can't reach the target position, or you should output "It takes n seconds to reach the target position, let me show you the way."(n is the minimum seconds), and tell our hero the whole path. Output a line contains "FINISH" after each test case. If there are more than one path, any one is OK in this problem. More details in the Sample Output.

 

Sample Input

5 6 .XX.1. ..X.2. 2...X. ...XX. XXXXX. 5 6 .XX.1. ..X.2. 2...X. ...XX. XXXXX1 5 6 .XX... ..XX1. 2...X. ...XX. XXXXX.

 

Sample Output

It takes 13 seconds to reach the target position, let me show you the way. 1s:(0,0)->(1,0) 2s:(1,0)->(1,1) 3s:(1,1)->(2,1) 4s:(2,1)->(2,2) 5s:(2,2)->(2,3) 6s:(2,3)->(1,3) 7s:(1,3)->(1,4) 8s:FIGHT AT (1,4) 9s:FIGHT AT (1,4) 10s:(1,4)->(1,5) 11s:(1,5)->(2,5) 12s:(2,5)->(3,5) 13s:(3,5)->(4,5) FINISH It takes 14 seconds to reach the target position, let me show you the way. 1s:(0,0)->(1,0) 2s:(1,0)->(1,1) 3s:(1,1)->(2,1) 4s:(2,1)->(2,2) 5s:(2,2)->(2,3) 6s:(2,3)->(1,3) 7s:(1,3)->(1,4) 8s:FIGHT AT (1,4) 9s:FIGHT AT (1,4) 10s:(1,4)->(1,5) 11s:(1,5)->(2,5) 12s:(2,5)->(3,5) 13s:(3,5)->(4,5) 14s:FIGHT AT (4,5) FINISH God please help our poor hero. FINISH