Sol-Dp-滑雪

Solution of SHOI2002-滑雪

Dp,注意要满足无后效性所以从大到小转移,用优先队列能过

demo:

#include<queue>
#include<iostream>
#include<cstdio>
using namespace std ;
struct node{
	int posx , posy ;
	int value ;
	bool operator<(node x) const {
		return this->value>x.value ;
	}
	node(int i,int j,int x):posx(i),posy(j),value(x){}
} ;
template<class T>
class comparer{
    public:
    	bool operator()(T x,T y){
			return x<y ;
		}
} ;
priority_queue<node,deque<node>,comparer<node> >q;
int n , m , maxn = -0x3f3f3f3f ;
int Dp[105][105] , mat[105][105];
#define read(X) scanf("%d",&X) 
#define loop(I,X) for(int I = 1 , end = (X) ; I <= end ; ++I)
int main(){
	read(n) , read(m) ;
	loop(i,n)
		loop(j,m)
		    Dp[i][j] = 1 , read(mat[i][j]) , q.push(node(i,j,mat[i][j])) ;
	while(!q.empty()){
		node temp = q.top() ; q.pop() ;
	    int i = temp.posx , j = temp.posy , val = temp.value ;
		if(mat[i-1][j] < val) Dp[i][j] = max(Dp[i][j] , Dp[i-1][j] + 1) ;
		if(mat[i+1][j] < val) Dp[i][j] = max(Dp[i][j] , Dp[i+1][j] + 1) ;
		if(mat[i][j-1] < val) Dp[i][j] = max(Dp[i][j] , Dp[i][j-1] + 1) ;
		if(mat[i][j+1] < val) Dp[i][j] = max(Dp[i][j] , Dp[i][j+1] + 1) ;
		if(maxn < Dp[i][j]) maxn = Dp[i][j] ;
	}
	printf("%d\n",maxn) ;
}

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转载自blog.csdn.net/qq_42000775/article/details/83500519