POJ 3278 第一道BFS 100題 紀念一下

題目在 http://poj.org/problem?id=3278

這個題目過的還是比較艱難的，主要是因為我沒有一下子用最好的辦法做，很多時候，還是沒有完全區分開什麼時候用BFS， 什麼時候用DFS

再做幾個題目慢慢總結吧，不過做了這個題目就有100道題目了，繼續加油呀~ Fight For oFFers

Source Code

Problem: 3278		User: hopeztm
Memory: 2308K		Time: 141MS
Language: C++		Result: Accepted

• Source Code

  #include <stdio.h>#include <memory.h>#include <queue>using namespace std;#define MAX_NUM 200003#define MAX_INT 1000000000#define MIN(x, y, z) (x < y ? x : y) < z ? (x < y ? x : y) : zint num[MAX_NUM];int visit[MAX_NUM];int From, To;int GetMinSteps(int i, int target){queue<int> t;t.push(i);int current;num[i] = 0;visit[i] = true;while(!t.empty()){current = t.front();t.pop();if(current == To)break;if( current - 1 >= 0 && !visit[current-1]){visit[current] = true;num[current - 1] = min(num[current - 1], num[current] + 1);t.push(current - 1);}if(current + 1 <= To && !visit[current + 1] ){visit[current + 1] = true;num[current+1] = min(num[current + 1], num[current] + 1);t.push(current + 1);}if(2 * current < 2 * To  && !visit[2 * current] ){visit[2 * current ] = true;num[current * 2] = min(num[current * 2], num[current] + 1);t.push(2 * current);}}return num[To];}int main(){while(scanf("%d%d", &From, &To) != EOF){memset(visit, 0, sizeof(visit));  memset(num, 0x7F, sizeof(num));printf("%d\n", GetMinSteps(From, To));}return 0;}