国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

hdu 1078 記憶化搜索

2019-11-11 06:16:31
字體:
來源:轉載
供稿:網友

FatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. At each grid location Fatmouse has hid between 0 and 100 blocks of cheese in a hole. Now he’s going to enjoy his favorite food.

FatMouse begins by standing at location (0,0). He eats up the cheese where he stands and then runs either horizontally or vertically to another location. The PRoblem is that there is a super Cat named Top Killer sitting near his hole, so each time he can run at most k locations to get into the hole before being caught by Top Killer. What is worse – after eating up the cheese at one location, FatMouse gets fatter. So in order to gain enough energy for his next run, he has to run to a location which have more blocks of cheese than those that were at the current hole.

Given n, k, and the number of blocks of cheese at each grid location, compute the maximum amount of cheese FatMouse can eat before being unable to move. Input There are several test cases. Each test case consists of

a line containing two integers between 1 and 100: n and k n lines, each with n numbers: the first line contains the number of blocks of cheese at locations (0,0) (0,1) … (0,n-1); the next line contains the number of blocks of cheese at locations (1,0), (1,1), … (1,n-1), and so on. The input ends with a pair of -1’s. Output For each test case output in a line the single integer giving the number of blocks of cheese collected. Sample Input 3 1 1 2 5 10 11 6 12 12 7 -1 -1 Sample Output 37

題意:在n*n的格子上,每個點各有若干塊奶酪,胖老鼠從左上角出發,每次最多走k步(只能直走),且下一點必須比這一點的奶酪多,問最多能吃到多少塊奶酪。 記憶化搜索 if(dp[x][y]) return; dp[x][y]=ans+v[xx][yy];

#include <bits/stdc++.h>using namespace std;int mp[2000][2000];int dp[2000][2000];int vis[2000][2000];int nex[4][2]={0,1,1,0,-1,0,0,-1};int n,k;int dfs(int x,int y){ int ans=0; if(!dp[x][y]) { for(int i=1;i<=k;i++){ for(int j=0;j<4;j++) { int xx=x+nex[j][0]*i; int yy=y+nex[j][1]*i; if(xx<1||xx>n||yy<1||yy>n||mp[xx][yy]<=mp[x][y]) continue; dfs(xx,yy); ans=max(ans,dp[xx][yy]); } } dp[x][y]=ans+mp[x][y]; } return dp[x][y];}int main(){ while(cin>>n>>k) { memset(dp,0,sizeof(dp)); if(n==-1&&k==-1) break; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) cin>>mp[i][j]; } printf("%d/n",dfs(1,1)); }}
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 奎屯市| 广东省| 沈阳市| 临澧县| 铁岭市| 宜州市| 祁东县| 石首市| 新和县| 三门县| 吴川市| 泰来县| 乡城县| 镇康县| 玛纳斯县| 吉林市| 天长市| 营口市| 沁源县| 策勒县| 长泰县| 涿鹿县| 平昌县| 漳浦县| 崇义县| 通州区| 启东市| 揭阳市| 唐山市| 镇赉县| 阳泉市| 衡水市| 蛟河市| 措美县| 天门市| 三河市| 潢川县| 彭阳县| 抚顺县| 库车县| 大英县|