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hdu---------(1026)Ignatius and the Princess I(bfs+dfs)，ignatiusbfs

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hdu---------(1026)Ignatius and the Princess I(bfs+dfs)，ignatiusbfs

Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12124 Accepted Submission(s): 3824
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

Author Ignatius.L 簡單的搜索題：記錄路徑額搜索... 代碼：

 1 /*Problem : 1026 ( Ignatius and the Princess I )     Judge Status : Accepted
 2 RunId : 11510650    Language : G++    Author : huifeidmeng
 3 Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta*/
 4 
 5 #include<cstring>
 6 #include<cstdio>
 7 #include<cstdlib>
 8 #include<queue>
 9 #include<iostream>
10 using namespace std;
11 const int maxn=101;
12 struct node{
13   int x,y;
14 };
15 struct sode{
16   int val;
17   int sum;
18   node pre; //他一個點
19   void setint(int x,int y){
20       pre.x=x;
21       pre.y=y;
22   }
23 };
24  int n,m,cont;
25  sode map[maxn][maxn];
26 int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
27 void bfs(){
28   queue<node> anc;
29   node tem={0,0};
30   anc.push(tem);
31   while(!anc.empty()){
32       node sav=anc.front();
33       anc.pop();
34     for(int i=0;i<4;i++){
35           tem=(node){dir[i][0]+sav.x,dir[i][1]+sav.y};
36         if(tem.x>=0&&tem.x<n&&tem.y>=0&&tem.y<m&&map[tem.x][tem.y].val!=-1){
37           if(map[tem.x][tem.y].sum==0
38           ||map[tem.x][tem.y].sum>map[sav.x][sav.y].sum+map[tem.x][tem.y].val+1)
39            {
40               map[tem.x][tem.y].sum=map[sav.x][sav.y].sum+map[tem.x][tem.y].val+1;
41               map[tem.x][tem.y].setint(sav.x,sav.y);
42               anc.push(tem);
43            }
44         }
45     }
46     }
47 }
48 void dfs(sode a,node cur){
49     if(cur.x==0&&cur.y==0){
50       cont=1;
51       while(map[cur.x][cur.y].val-->0)
52        printf("%ds:FIGHT AT (%d,%d)\n",cont++,cur.x,cur.y);
53      //  printf("%ds:(%d,%d)->(%d,%d)\n",cont++,a.pre.x,a.pre.y,cur.x,cur.y);
54     return;
55     }
56     dfs(map[a.pre.x][a.pre.y],a.pre);
57     printf("%ds:(%d,%d)->(%d,%d)\n",cont++,a.pre.x,a.pre.y,cur.x,cur.y);
58     if(a.val>0){
59       while(a.val-->0){
60          printf("%ds:FIGHT AT (%d,%d)\n",cont++,cur.x,cur.y);
61      }
62     }
63 }
64 int main(){
65   char ss[105];
66   //freopen("test.in","r",stdin);
67  // freopen("test.out","w",stdout);
68  while(scanf("%d%d",&n,&m)!=EOF){
69    for(int i=0;i<n;i++){
70      scanf("%s",ss);
71     for(int j=0;j<m;j++){
72         map[i][j].sum=0;
73         map[i][j].setint(0,0);
74        if(ss[j]=='.') map[i][j].val=0;
75        else if(ss[j]=='X')map[i][j].val=-1;
76        else{
77             map[i][j].val= ss[j]-'0';
78             if(i+j==0)map[i][j].sum=map[i][j].val;
79        }
80      }
81 }
82   bfs();
83 if(map[n-1][m-1].pre.x==0)
84    printf("God please help our poor hero.\n");
85 else {
86     printf("It takes %d seconds to reach the target position, let me show you the way.\n",map[n-1][m-1].sum);
87     dfs(map[n-1][m-1],(node){n-1,m-1});
88 }
89   puts("FINISH");
90 }
91  return 0;
92 }

HDU 1026 Ignatius and the Princess I 哪位大牛可以幫幫小菜改一下這個代碼？一直MLE

親，你的想法和我最開始做這個題的時候是一樣的，先bfs找到耗時最小值，後dfs找到路徑。
看了你的代碼，我提醒幾點（這是很久的帖子了，也許你已經成了大牛了，我隨便說的，莫要見怪啊）
（1）大牛們說了，沒有注釋的代碼就是垃圾，我表示你的代碼我咋的一看真沒看懂，自己烤走稍作注釋後才看懂你的思路。以下是我做注釋後你的代碼：
#include<iostream>#include<queue>using namespace std;int dirt[4][2]= {{1,0},{-1,0},{0,1},{0,-1}}; //四個方向struct nobe //存儲地圖情況，坐標，耗時，地圖輸入情況，是否訪問{ int x,y,time,data,v;};nobe a[500]; //存儲路徑情況nobe map[101][101]; ////存儲地圖情況int l,n,m;void bfs() //廣度優先搜索{ nobe first,next; queue <nobe> Q; int i; first.x=0; first.y=0; first.time=0; Q.push(first); //起點入隊 map[0][0].time=0; while(!Q.empty()) { first=Q.front(); //彈出隊頭節點 Q.pop(); for(i=0; i<4; i++) //四個方向 { next.x=first.x+dirt[i][0]; next.y=first.y+dirt[i][1]; if(next.x>=0&&next.x<n&&next.y>=0&&next.y<m&&map[next.x][next.y].data!=-1&&first.time+map[next.x][next.y].data<map[next.x][next.y].time) { //節點在地圖內，該點處不是牆，該點耗時最優 next.time=first.time+1; //耗時增加 next.time=next.time+map[next.x][next.y].data; //如果要打怪，就把時間加上 map[next.x][next.y].time=next.time; //給地圖數組中的時間屬性更新至最優值 Q.push(next); //新節點入隊 } } }}void savetu(int d){ int i,j; nobe first,next; queue <nobe> Q; for(i=0; i<d; i++) //初始化為最大值 a......余下全文>>