POJ 1088 比较有意思的动态规划

注意2种方法的差别,递归的使用

#include<iostream>using namespace std;int r,c;int xd[101][101];int len[101][101];//上下左右int pos[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};bool ok(int i,int j){if(1 <= i && i <= r && 1 <= j && j <= c)return true;return false;}int dp(int i,int j){if(len[i][j] > 0)return len[i][j];/*int up,down,left,right,maxUD,maxLR,max;up = down = left = right = 0;if(i - 1 >= 1 && xd[i - 1][j] < xd[i][j])up = dp(i - 1,j) + 1;if(i + 1 <= r && xd[i + 1][j] < xd[i][j])down = dp(i + 1,j) + 1;//当2者都为0的时候照样成立maxUD = up > down ? up : down;if(j - 1 >= 1 && xd[i][j - 1] < xd[i][j])left = dp(i,j - 1) + 1;if(j + 1 <= c && xd[i][j + 1] < xd[i][j])right = dp(i,j + 1) + 1;maxLR = left > right ? left : right;max = maxLR > maxUD ? maxLR : maxUD;len[i][j] = max;return max;*///method2int k;for(k = 0;k < 4;k++)if(ok(i + pos[k][0],j + pos[k][1]))if(xd[i][j] > xd[i + pos[k][0]][j + pos[k][1]])if(len[i][j] < dp(i + pos[k][0],j + pos[k][1]) + 1)len[i][j] = dp(i + pos[k][0],j + pos[k][1]) + 1;return len[i][j];}int main(){cin>>r>>c;int i,j;for(i = 1;i <= r;i++)for(j = 1;j <= c;j++){cin>>xd[i][j];len[i][j] = 0;}int max = 0;for(i = 1;i <= r;i++){for(j = 1;j <= c;j++){if(max < dp(i,j))max = dp(i,j);cout<<len[i][j]<<" ";}cout<<endl;}cout<<max + 1<<endl;return 0;}