【UVA】10285-Longest Run on a Snowboard（動態規劃）

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這題出簡單了，不需要列印路徑。

狀態方程dp[i][j] = max(dp[i-1][j],dp[i][j-1],dp[i+1][j],dp[i][j+1]);

14003395

10285

Longest Run on a Snowboard

Accepted

C++

0.026

2014-08-07 11:43:51

枚舉每個點進行遍曆，使用記憶化搜尋。要不會逾時。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<stack>#include<queue>#include<map>#include<set>#include<list>#include<string>#include<sstream>#include<ctime>using namespace std;#define _PI acos(-1.0)#define INF (1 << 10)#define esp 1e-6typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> pill;/*======================================================================================*/#define MAXD 100 + 10char name[MAXD];int m,n,ans;int dp[MAXD][MAXD];int mat[MAXD][MAXD];int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};int DP(int x,int y){    if(dp[x][y] != -1)        return dp[x][y];    dp[x][y] = 1;    for(int i = 0 ; i < 4 ; i++){        int _x = dir[i][0] + x;        int _y = dir[i][1] + y;        if(_x >= 0 && _y >=0 && _x < n && _y < m && mat[x][y] < mat[_x][_y]){            dp[x][y] = max(dp[x][y],DP(_x,_y) + 1);        }    }    ans = max(ans,dp[x][y]);    return dp[x][y];}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%s%d%d",name,&n,&m);        int pos_x,pos_y;        int MIN = INF;        memset(dp,-1,sizeof(dp));        for(int i = 0 ; i < n ; i++)            for(int j = 0 ; j < m ; j++){                scanf("%d",&mat[i][j]);                if(mat[i][j] < MIN){                    MIN = mat[i][j];                    pos_x = i;                    pos_y = j;                }            }        ans = 0;        for(int i = 0 ; i < n ; i++)            for(int j = 0 ; j < m ; j++)                DP(i,j);        printf("%s: %d\n",name,ans);    }    return 0;}