小马的博客

#include#include#include#includeusing namespace std;char map[25][25];int PRE[25][25];int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};int m,n;int sx,sy,ex,ey;struct node{ int x; int y;

BFS+同于定理。（看了解题报告才会的），这题实在太巧妙#include#include#includeusing namespace std;const int maxm=1e4+10;int vis[maxm];int d[maxm];int m,n;int ok;struct node{ int pre; int yu; int dig;}que

题意：求最小的线段树的右端点（根节点表示区间[0,n]），使得给定的区间[L,R]是线段树的某个节点。#include#include#includeusing namespace std;#define LL long longconst LL inf=0x3f3f3f3f3f3f3f3f;LL ans;void dfs(LL l,LL r){ if(l==0)

FriendsTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1946    Accepted Submission(s): 975Problem DescriptionThere are n people

乘法游戏是在一行牌上进行的。每一张牌包括了一个正整数。在每一个移动中，玩家拿出一张牌，得分是用它的数字乘以它左边和右边的数，所以不允许拿第1张和最后1张牌。最后一次移动后，这里只剩下两张牌。         你的目标是使得分的和最小。例如，如果数是10  1  50  20  5，依次拿1、20、50，总分是10*1*50+50*20*5+10*50*5=8000而拿50、20、1，总分是

题意：给出一个长l的木棍， 再给出n个要求切割的点，每次切割的代价是当前木棍的长度， 现在要求输出最小代价。/*给定一段len长的木棍，有n个切割点，每个切割点切掉的花费是当前切割点所在木棍的长度，求最少的花费。*/#include#include#include#include#include#include#includeusing namespace std;const

Prime PathTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16065 Accepted: 9068DescriptionThe ministers of the cabinet were quite upset by the message fr

DessertTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 5473 Accepted: 2048DescriptionFJ has a new rule about the cows lining up for dinner. Not only m

A. Oleg and sharestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputOleg the bank client checks share prices

基础BFS#include#include#include#include#includeusing namespace std;const int maxm=205;const int inf=1<<29;int m,n;int x1,y1,x2,y2;char map[maxm][maxm];int vis[maxm][maxm];int w[maxm][maxm];

就是用DFS找个数#include#include#includeusing namespace std;const int maxm=55;char map[maxm][maxm];int DIR[22][2] = {{0,1},{1,0},{0,-1},{-1,0}};int dir[22][4] = {{0,0,1,1}, {1,0,0,1}, {0,1,1,0}, {1,1,

题意：判断，能否从S到D，恰好在T时间到。#include#include#includeusing namespace std;const int maxm=10;int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};char map[maxm][maxm];int vis[maxm][maxm];int x1,y1;int m,n,T;in

#include#include#include#includeusing namespace std;const int maxm=150;int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};char s[maxm][maxm];int map[maxm][maxm];int vis[maxm][maxm];int list[maxm][max

暴力水过#include#include#includeusing namespace std;int main(){ int n; while(scanf("%d",&n)!=EOF) { int a,b,c,d; for(a=2;a<=n;a++) { for(b=2;b<a;b++)

坑点：1：当两边都是#号是，是死路            2：当两边种，一边是#，另一边是*是，也是死路。其他都没啥好说的这是一道比较难搞定的搜索题，三维立体搜索。#include#include#includeusing namespace std;const int maxm=20;char map[maxm][maxm][maxm];int vis[maxm][max

很简单的三维深搜。#include#include#includeusing namespace std;#define N 60int map[N][N][N];int m,n,k,time,ok;int a1,b1,c1,a2,b2,c2;int dir[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};st

简单的搜索题#include#include#includeusing namespace std;#define N 30int vis[N][N];int t[N];int num[N];int k;int p;void dfs(int temp,int cur){ if(cur==20&&vis[num[19]][k]) { printf

#include#include#includeusing namespace std;const int maxm=1e4+10;const int inf=1<<29;int mins[maxm],minv[maxm];int n,m;int bests;void dfs(int s,int v,int level,int r,int h){ if(level==0

读懂题意就知道是一个简单的DFS了#include#include#includeusing namespace std;#define N 30int vis[N][N];int a[N][N];int m,n;int ok;int dir[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};vo

BFS题，so easy!#include#include#include#includeusing namespace std;const int maxm=10;int vis[maxm][maxm][maxm][maxm];int m,n;struct node{ int a[4]; int step;}num1,num2;void bfs(){

胜利大逃亡(续)Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9153    Accepted Submission(s): 3303Problem DescriptionIgnatius再次被魔王抓走了(