动态规划(二）

Posted 2020-08-30

                            动态规划（二）

uva437 巴比伦塔

题目链接：https://vjudge.net/problem/UVA-437

题意：这题可以看做是三维的DAG问题。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 92
int dp[maxn], dps[maxn];
int G[maxn][maxn];
int k;
int x[maxn], y[maxn];

int dp_find(int i)
{
    int& ans = dps[i];
    if (ans > 0) return ans;
    ans = dp[i];
    for (int j = 0; j < k; j++){
        if (G[i][j]){
            ans = max(ans, dp_find(j) + dp[i]);
        }
    }
    return ans;
}

int main()
{
    int T, kase;
    kase = 0;
    while (scanf("%d",&T)!=EOF&&T){
        int a[3];
        int  ans; k = 0;
        for (int j = 0; j < T; j++){
            cin >> a[0] >> a[1] >> a[2];
            for (int i = 0; i < 3; i++){
                dp[k] = a[i];
                if (a[(i + 1) % 3] < a[(i + 2) % 3]){
                    x[k] = a[(i + 1) % 3];
                    y[k] = a[(i + 2) % 3];
                }
                else{
                    x[k] = a[(i + 2) % 3];
                    y[k] = a[(i + 1) % 3];
                }
                k++;
            }
        }
        memset(G, 0, sizeof(G));
        for (int i = 0; i < k;i++)
            for (int j = 0; j < k; j++){
                if (x[i] < x[j] && y[i] < y[j]){
                    
                    G[i][j] = 1;
                }
        }
        memset(dps, 0, sizeof(dps));
        ans = 0;

        for (int i = 0; i < k; i++){
            if (dp_find(i)>ans)
                ans = dp_find(i);
        }
        printf("Case %d: maximum height = %d\n", ++kase, ans);
    }
    return 0;
}