poj 1163 The Triangle 搜索 难度:0

 

The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 37931		Accepted: 22779

Description

7
3   8
8   1   0
2   7   4   4
4   5   2   6   5

(Figure 1)

Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right. 

Input

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output

Your program is to write to standard output. The highest sum is written as an integer.

Sample Input

5
7
3 8
8 1 0 
2 7 4 4
4 5 2 6 5

Sample Output

30
记忆化搜索

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=211;
int  num[maxn][maxn];
int n,m;
int maz[maxn][maxn];
const int dx[2]={1,1},dy[2]={0,1};
int dfs(int sx,int sy){
    if(num[sx][sy]!=-1)return num[sx][sy];
    int tmp=maz[sx][sy];
    for(int i=0;i<2;i++){
        int tx=sx+dx[i],ty=sy+dy[i];
        if(tx>=0&&tx<n&&ty>=0&&ty<tx+1){
            int t=dfs(tx,ty);
            tmp=max(t+maz[sx][sy],tmp);
        }
    }
    return num[sx][sy]=tmp;
}
int main(){
    int T=1;
    while(T--&&scanf("%d",&n)==1){
        for(int i=0;i<n;i++){
            for(int j=0;j<i+1;j++){
                scanf("%d",maz[i]+j);
            }
        }
        memset(num,-1,sizeof(num));
        int ans=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<i+1;j++){
                if(num[i][j]==-1){
                    int tmp=dfs(i,j);
                    ans=max(ans,tmp);
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}