标签：uva 10285 the tower 记忆化搜索 dfs
Problem C
Longest Run on a Snowboard
Input: standard input
Output: standard output
Time Limit: 5 seconds
Memory Limit: 32 MB
Michael likes snowboarding. That‘s not very surprising, since snowboarding is really great. The bad thing is that in order to gain speed, the area must slide downwards. Another disadvantage is that when you‘ve reached the bottom of the hill you have to walk up again or wait for the ski-lift.
Michael would like to know how long the longest run in an area is. That area is given by a grid of numbers, defining the heights at those points. Look at this example:
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
One can slide down from one point to a connected other one if and only if the height decreases. One point is connected to another if it‘s at left, at right, above or below it. In the sample map, a possible slide would be 24-17-16-1 (start at 24, end at 1). Of course if you would go 25-24-23-...-3-2-1, it would be a much longer run. In fact, it‘s the longest possible.
Input
The first line contains the number of test cases N. Each test case starts with a line containing the name (it‘s a single string), the number of rows R and the number of columns C. After that follow R lines with C numbers each, defining the heights. R and C won‘t be bigger than 100, N not bigger than 15 and the heights are always in the range from 0 to 100.
For each test case, print a line containing the name of the area, a colon, a space and the length of the longest run one can slide down in that area.
Sample Input
2
Feldberg 10 5
56 14 51 58 88
26 94 24 39 41
24 16 8 51 51
76 72 77 43 10
38 50 59 84 81
5 23 37 71 77
96 10 93 53 82
94 15 96 69 9
74 0 62 38 96
37 54 55 82 38
Spiral 5 5
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
Sample Output
Feldberg: 7
Spiral: 25
(Math Lovers’ Contest, Problem Setter: Stefan Pochmann)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;
const int INF = 1000010;
using namespace std;
int mp[110][110],dp[110][110];
int n,m;
char s[140];
int xx[4]= {-1,0,1,0};
int yy[4]= {0,1,0,-1};
bool vis[110][110];
int dfs(int x,int y)
{
if(dp[x][y]!=1)
return dp[x][y];
int ans=1;
for(int i=0; i<4; i++)
{
int nx=xx[i]+x;
int ny=yy[i]+y;
if(nx<0||nx>=n||ny<0||ny>=m||mp[nx][ny]>=mp[x][y])
continue;
dp[x][y]=max(dp[x][y],dfs(nx,ny)+1);
}
return dp[x][y];
}
int main()
{
int t;
while(~scanf("%d",&t))
{
while(t--)
{
scanf("%s",s);
scanf("%d%d",&n,&m);
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
scanf("%d",&mp[i][j]);
int ans=1;
memset(vis,0,sizeof vis);
for(int i=0;i<=n+1;i++)
fill(dp[i],dp[i]+m+1,1);
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
ans=max(ans,dfs(i,j));
printf("%s: %d\n",s,ans);
}
}
return 0;
}
uva 10285 The Tower of Babylon（记忆化搜索）
标签：uva 10285 the tower 记忆化搜索 dfs
原文地址：http://blog.csdn.net/acm_baihuzi/article/details/43054847
