HDU1242:Rescue bfs和dfs
2015-08-04 16:00
369 查看
Description
Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.
Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.
You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)
Input
First line contains two integers stand for N and M.
Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.
Process to the end of the file.
Output
For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."
Sample Input
7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........
Sample Output
13
题意:求从r到a的最短路径。x表示要走两步。
这题用dfs和bfs都能写。个人建议用bfs,因为dfs要剪枝,太麻烦了。不管用什么方法都应从a搜到r,因为r可能有多个。
dfs
dfs,上面同学看不懂,就加了下面这种,哎,写着么好,居然看不懂
Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.
Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.
You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)
Input
First line contains two integers stand for N and M.
Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.
Process to the end of the file.
Output
For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."
Sample Input
7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........
Sample Output
13
题意:求从r到a的最短路径。x表示要走两步。
这题用dfs和bfs都能写。个人建议用bfs,因为dfs要剪枝,太麻烦了。不管用什么方法都应从a搜到r,因为r可能有多个。
dfs
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n,m,mn; int v[205][205]; char map[205][205]; int yi[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; void dfs(int x,int y,int k) { if (k>=mn) return ; int i,j,rx,ry; for (i=0;i<4;i++) { rx=x+yi[i][0]; ry=y+yi[i][1]; if (map[rx][ry]=='r') { mn=min(mn,k+1); return ; } if (rx>=1&&rx<=n&&ry>=1&&ry<=m&&map[rx][ry]!='#'&&!v[rx][ry]) { v[rx][ry]=1; if (map[rx][ry]=='x') dfs(rx,ry,k+2); else dfs(rx,ry,k+1); v[rx][ry]=0; } } } int main() { int i,j,x,y; while (~scanf("%d%d",&n,&m)) { for (i=1;i<=n;i++) for (j=1;j<=m;j++) { scanf(" %c",&map[i][j]); if (map[i][j]=='a') { x=i; y=j; } } mn=100000; memset(v,0,sizeof(v)); dfs(x,y,0); if (mn==100000) printf("Poor ANGEL has to stay in the prison all his life.\n"); else printf("%d\n",mn); } } bfs
#include<cstdio> #include<cstring> #include<algorithm> #include<queue> using namespace std; int n,m,ans; char map[205][205]; int yi[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; struct f { int x,y; int t; };f s,r; int bfs(int x,int y) { int i,j; queue<f>q; s.x=x; s.y=y; s.t=0; map[x][y]='#'; q.push(s); while (!q.empty()) { s=q.front(); q.pop(); for (i=0;i<4;i++) { r.x=s.x+yi[i][0]; r.y=s.y+yi[i][1]; if (map[r.x][r.y]=='r') return s.t+1; if (r.x>=0 && r.x<n && r.y>=0 && r.y<m && map[r.x][r.y]!='#') { if (map[r.x][r.y]=='x') r.t=s.t+2; else r.t=s.t+1; map[r.x][r.y]='#'; q.push(r); } } } return -1; } int main() { int i,j,x,y; while (~scanf("%d%d",&n,&m)) { if (n==0&&m==0) break; ans=0; for (i=0;i<n;i++) for (j=0;j<m;j++) { scanf(" %c",&map[i][j]); if (map[i][j]=='a') {x=i;y=j;} } ans=bfs(x,y); if (ans==-1) printf("Poor ANGEL has to stay in the prison all his life.\n"); else printf("%d\n",ans); } return 0; }
dfs,上面同学看不懂,就加了下面这种,哎,写着么好,居然看不懂
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n,m,mn; int v[205][205]; int v1[205][205]; char map[205][205]; int yi[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; void dfs(int x,int y,int k) { if (x<1||x>n||y<1||y>m||map[x][y]=='#'||v[x][y]) return ; if (k>=mn) return ; //剪枝,加这句怎么增加了30多ms if (v1[x][y]&&k>=v1[x][y]) return ;//剪枝,加这句能减少60多ms. else v1[x][y]=k; // if (map[x][y]=='r') mn=k; v[x][y]=1; if (map[x][y]=='x') k+=2; else k++; dfs(x+1,y,k); dfs(x-1,y,k); dfs(x,y+1,k); dfs(x,y-1,k); v[x][y]=0; } int main() { int i,j,x,y; while (~scanf("%d%d",&n,&m)) { for (i=1;i<=n;i++) for (j=1;j<=m;j++) { scanf(" %c",&map[i][j]); if (map[i][j]=='a') { x=i; y=j; } } mn=100000; memset(v,0,sizeof(v)); memset(v1,0,sizeof(v1)); dfs(x,y,0); if (mn==100000) printf("Poor ANGEL has to stay in the prison all his life.\n"); else printf("%d\n",mn); } }
相关文章推荐
- EasyUI——常见用法总结
- (一〇六)iPad开发之UIPopoverController的使用
- (一〇六)iPad开发之UIPopoverController的使用
- iOS 并发编程之 Operation Queues
- UINavigationBar-使用总结
- StringBuffer跟StringBuilder的区别
- UIModalPresentationStyle和UIModalTransitionStyle
- EazyUI Grid和TreeGrid的 CheckBox 模式
- UIlabel文字大小自适应label宽度变大变小
- android多分辨率多屏幕密度下UI适配方案
- iOS编程——简单的UI自动适配解决方案:Masonry
- SQL server字段名中有数据类型为text时,adoquery->fieldbyname->asstring取不到全部值
- UITools
- UI_Stepper
- IOS开发之关于UIButton点击没有响应问题
- uva 10706 Number Sequence(数学)
- [置顶] 项目中requireJS的应用例子
- perf_events Frequently Asked Questions
- iOS学习笔记6- iOS UICollectionView简单使用
- The method build() is undefined for the type Notification.Builder