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 04:50:22
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: 沿河| 灵丘县| 翼城县| 沁阳市| 察隅县| 贵南县| 天峻县| 颍上县| 凉城县| 建宁县| 九寨沟县| 青田县| 罗田县| 措勤县| 海口市| 连城县| 郎溪县| 南皮县| 崇文区| 宁德市| 怀远县| 山西省| 玉门市| 诏安县| 江津市| 蒙山县| 德化县| 石棉县| 忻州市| 易门县| 徐汇区| 华宁县| 凭祥市| 博罗县| 永定县| 永修县| 曲周县| 靖州| 广东省| 平南县| 金溪县|