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Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 88861 | Accepted: 33344 |
Description
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Input
Output
Sample Input
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Sample Output
25
很简单的题目,一遍过,因为题目要求的是从大到小的路径,所以根本就不需要用vist标记已经访问的节点了,这条路本身就是自带有向属性
#include <iostream> #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <string> #include <queue> #include <stack> #include <algorithm> const int inf = (1<<31)-1; const int MAXN = 1e2+10; using namespace std; int dp[MAXN][MAXN]; int vist[MAXN][MAXN]; int a[MAXN][MAXN]; int n,m; int mov[4][2]={-1,0,1,0,0,1,0,-1}; int check(int x,int y){ if(x<1||y<1||x>n||y>m)return 0; // if(vist[x][y]==1)return 0; else return 1; } void dfs(int x,int y){ int nx,ny; // cout<<"haha"<<endl; for(int i=0;i<4;i++){ nx = x+mov[i][0]; ny = y+mov[i][1]; if(check(nx,ny)&&a[x][y]>a[nx][ny]){ if(dp[nx][ny]==1){ // vist[nx][ny] = 1; dfs(nx,ny); dp[x][y] = max(dp[x][y],dp[nx][ny]+1); //vist[nx][ny] = 0; } else{ dp[x][y] = max(dp[x][y],dp[nx][ny]+1); } } } } int main() { while(scanf("%d%d",&n,&m)!=EOF){ int mmax = -inf; for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ scanf("%d",&a[i][j]); dp[i][j] = 1; vist[i][j] = 0; } } for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(dp[i][j]!=1||vist[i][j]==0){ //if() vist[i][j] = 1; dfs(i,j); } mmax = max(mmax,dp[i][j]); } } /*for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ printf("%d ",dp[i][j]); } cout<<endl; }*/ cout<<mmax<<endl; } //cout << "Hello world!" << endl; return 0; }
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原文地址:http://www.cnblogs.com/EdsonLin/p/5441033.html