book[n]=1;intnext[3]={1,-1,2};intflag=0;while(head<tail){for(inti=0;i<3;i++){intdx;if(i!=2) dx=s[head].x+next[i];elsedx=s[head].x*next[i];if(dx<0||dx>400005)continue;if(book[dx]==0){ book[dx]=1; s[tail].x=dx; s[tail++].s=s[head].s+1; }if(dx=...
queue<node> q; node s; s.d=n; s.time=0; q.push(s); vis[n]=1;while(!q.empty()) { node t=q.front();if(t.d==k) { cout<<t.time<<endl;break; } q.pop();intflag=0;for(inti=0; i<3; i++) {if(i==0&&t.d+1<maxn&&!vis[t.d+1]) {nodea(t.d+1,t.time+1...
【POJ 3278】Catch That Cow 题解(广度优先搜索) 农夫约翰已被告知一头逃亡奶牛的位置,并想立即抓住她。他从一条数字线上的N点(0≤N≤100000)开始,奶牛在同一条数字线上的K点(0≥K≤100000)。农夫约翰有两种交通方式:步行和坐车。 *步行:FJ可以在一分钟内从任何点X移动到点X-1或X+1 *坐车:FJ可以在一...
【POJ 3278】Catch That Cow 题解(深度优先搜索) 农夫约翰已被告知一头逃亡奶牛的位置,并想立即抓住她。他从一条数字线上的N点(0≤N≤100000)开始,奶牛在同一条数字线上的K点(0≥K≤100000)。农夫约翰有两种交通方式:步行和坐车。 *步行:FJ可以在一分钟内从任何点X移动到点X-1或X+1 *坐车:FJ可以在一...
POJ 3278 Catch That Cow题目有哪些陷阱? Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 88732 Accepted: 27795 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,...
抓住那头牛(POJ3278) 题目: https://cn.vjudge.net/problem/POJ-3278 代码如下: 广度优先搜索(Breadth First Search)过程: (1) 把初始节点S0放入Open表中; (2) 如果Open表为空,则问题无解,失败退出; (3) 把Open表的第一个节点取出放入 Closed表,并记该节点为n; (4) 考察节点n是否为目标节点。若是...
poj3278 题⽬名称：Catch That Cow 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 ...
【poj3278 Catch That Cow---bfs】这个 代码混乱。。好歹过了 #include <iostream> #include <cstring> #include <string> #include <cstdio>...http://t.cn/8kM2Msa
(2)广度优先搜索(poj3278,poj1426,poj3126,poj3087.poj3414) (3)简单搜索技巧和剪枝(poj2531,poj1416,poj2676,1129) 五.动态规划 (1)背包问题. (poj1837,poj1276) (2)型如下表的简单DP(可参考lrj的书 page149): 1.E[j]=opt{D[i]+w(i,j)} (poj3267,poj1836,poj1260,poj2533) ...