UVA 10285 Longest Run on a Snowboard
2016-08-18 22:47
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题目链接:http://acm.hust.edu.cn/vjudge/problem/19213
题意:在n*n的格子里,每个格子有一个数,起点任意,每次只能移动到比当前格子数小的格子,问最长路径。
思路:f[i][j]表示从(i,j)为起点的最长路,采用记忆化搜索向四个方向搜索并取最大值。
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <utility>
using namespace std;
#define rep(i,j,k) for (int i=j;i<=k;i++)
#define Rrep(i,j,k) for (int i=j;i>=k;i--)
#define Clean(x,y) memset(x,y,sizeof(x))
#define LL long long
#define ULL unsigned long long
#define inf 0x7fffffff
#define mod 100000007
const int dx[] = {0,0,1,-1};
const int dy[] = {1,-1,0,0};
const int maxn = 200;
int f[maxn][maxn];
int a[maxn][maxn];
int n,m;
string name;
void init()
{
rep(i,1,n)
rep(j,1,m)
scanf("%d",&a[i][j]);
Clean(f,-1);
}
int cal( int x ,int y )
{
if ( f[x][y] != -1 ) return f[x][y];
int ans = 1;
rep(i,0,3)
{
int tx = x + dx[i];
int ty = y + dy[i];
if ( tx >= 1 && tx <= n && ty >= 1 && ty <= m && a[x][y] > a[tx][ty] )
ans = max( ans , cal( tx , ty ) + 1 );
}
return f[x][y] = ans;
}
void solve()
{
int ans = 0;
rep(i,1,n)
rep(j,1,m)
ans = max( ans , cal( i , j ) );
cout<<name<<": "<<ans<<endl;
}
int main()
{
int T;
cin>>T;
while( T-- )
{
cin>>name>>n>>m;
init();
solve();
}
return 0;
}
题意:在n*n的格子里,每个格子有一个数,起点任意,每次只能移动到比当前格子数小的格子,问最长路径。
思路:f[i][j]表示从(i,j)为起点的最长路,采用记忆化搜索向四个方向搜索并取最大值。
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <utility>
using namespace std;
#define rep(i,j,k) for (int i=j;i<=k;i++)
#define Rrep(i,j,k) for (int i=j;i>=k;i--)
#define Clean(x,y) memset(x,y,sizeof(x))
#define LL long long
#define ULL unsigned long long
#define inf 0x7fffffff
#define mod 100000007
const int dx[] = {0,0,1,-1};
const int dy[] = {1,-1,0,0};
const int maxn = 200;
int f[maxn][maxn];
int a[maxn][maxn];
int n,m;
string name;
void init()
{
rep(i,1,n)
rep(j,1,m)
scanf("%d",&a[i][j]);
Clean(f,-1);
}
int cal( int x ,int y )
{
if ( f[x][y] != -1 ) return f[x][y];
int ans = 1;
rep(i,0,3)
{
int tx = x + dx[i];
int ty = y + dy[i];
if ( tx >= 1 && tx <= n && ty >= 1 && ty <= m && a[x][y] > a[tx][ty] )
ans = max( ans , cal( tx , ty ) + 1 );
}
return f[x][y] = ans;
}
void solve()
{
int ans = 0;
rep(i,1,n)
rep(j,1,m)
ans = max( ans , cal( i , j ) );
cout<<name<<": "<<ans<<endl;
}
int main()
{
int T;
cin>>T;
while( T-- )
{
cin>>name>>n>>m;
init();
solve();
}
return 0;
}
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