标签：bfs
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 52335 | Accepted: 16416 |
Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Output
Sample Input
5 17
Sample Output
4
Hint
简单bfs， 每个状态可以转换成其他三个状态枚举就好了。
原来是打算用spfa， 但是TL了
代码：
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <vector>
const int M = 1e5+50;
using namespace std;
struct node{
int x;
int step;
bool operator < (const node &a) const {
return step > a.step;
}
};
bool vis[M];
int bfs(int n, int k){
memset(vis, 0, sizeof(vis));
if(n == k) return 0;
node st;
st.x = n; st.step = 0;
priority_queue<node> q;
q.push(st);
while(!q.empty()){
node temp = q.top();
q.pop();
node cur;
cur.step = temp.step+1;
cur.x = temp.x+1;
if(cur.x == k) return cur.step;
if(cur.x < M&&!vis[cur.x]){
vis[cur.x] = 1;
q.push(cur);
}
cur.x = temp.x-1;
if(cur.x == k) return cur.step;
if(cur.x > 0&&cur.x < M&&!vis[cur.x]){
vis[cur.x] = 1;
q.push(cur);
}
cur.x = temp.x*2;
if(cur.x == k) return cur.step;
if(cur.x < M&&!vis[cur.x]){
vis[cur.x] = 1;
q.push(cur);
}
}
}
int main(){
int n, k;
while(scanf("%d%d", &n, &k) == 2){
printf("%d\n", bfs(n, k));
}
}
/*vector <int >map[M*4];
int d[M];
bool vis[M];
/*int bfs(int n, int k){
memset(vis, 0, sizeof(vis));
priority_queue<node >q;
node st;
st.x = n; st.step = 0;
vis[n] = 1;
q.push(st);
if(n == k) return 0;
while(!q.empty()){
node temp = q.top();
q.pop();
node cur;
cur.step = temp.step+1;
cur.x = temp.x+1;
if(cur.x == k) return cur.step;
if(cur.x > 0&&cur.x < M){
vis[cur.x] = 1; q.push(cur);
}
cur.x = temp.x-1;
if(cur.x == k) return cur.step;
if(cur.x > 0&&cur.x < M){
vis[cur.x] = 1; q.push(cur);
}
cur.x = temp.x*2;
if(cur.x == k) return cur.step;
if(cur.x > 0&&cur.x < M){
vis[cur.x] = 1; q.push(cur);
}
}
}*/
/*void spfa(int n, int k){
memset(vis, 0, sizeof(vis));
memset(d, 'a', sizeof(d));
queue<int >q;
q.push(n);
d[n] = 0;
vis[n] = 1;
while(!q.empty()){
int temp = q.front();
q.pop();
for(int i = 0; i < map[temp].size(); ++ i){
if(d[map[temp][i]] > d[temp]+1){
d[map[temp][i]] = d[temp]+1;
if(!vis[map[temp][i]]){
vis[map[temp][i]] = 1;
q.push(map[temp][i]);
}
}
}
}
}
int main(){
int n, k;
while(scanf("%d%d", &n, &k) == 2){
for(int i = 1; i < M; ++ i){
map[i].clear();
if(i-1 > 0) map[i].push_back(i-1);
if(i+1 < M) map[i].push_back(i+1);
if(i*2 < M) map[i].push_back(i*2);
}
spfa(n, k);
//for(int i = 0; i < M; ++ i) vis[i] = 0, map[i].clear();
//for()
printf("%d\n", d[k]);
}
return 0;
}*/标签：bfs
原文地址：http://blog.csdn.net/shengweisong/article/details/44425693
