UVa 437 the Tower of Babylon:dp&dag

Source: Internet
Author: User
Tags cas time limit

437-the Tower of Babylon

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_ problem&problem=378


For a (x,y,z) brick, it can be placed in 3 positions: (top two for ground, rear one for high)

X, Y, Z

X, Z, y

Y, Z, x

Record each posture and become a brick with a 3*n fixed posture.

Then build the map and find the longest road on the DAG.

Complete code:

/*0.015s*/#include <bits/stdc++.h> using namespace std;  
int a[100][3], dp[100], N;  
    int DAG (int x) {if (dp[x]!=-1) return dp[x];  
        DP[X] = a[x][2];///high for (int y = 0; y < n; ++y) {if (y = = x) continue; if (A[y][0] < a[x][0] && a[y][1] < a[x][1] && Dp[x] < DAG (y) + a[x][2]) dp[x] = DAG  
    (y) + a[x][2];  
return dp[x];  
    int main () {int cas = 0, maxh, I;  
        while (scanf ("%d", &n), n) {n + = n << 1;  
        Memset (DP,-1, sizeof (DP));  
            for (i = 0; i < n; i + + 3) {scanf ("%d%d%d", &a[i][0], &a[i][1], &a[i][2));  
            Sort (A[i], A[i] + 3);  
            A[i + 1][0] = a[i][0], a[i + 1][1] = a[i][2], a[i + 1][2] = a[i][1];  
        A[i + 2][0] = a[i][1], a[i + 2][1] = a[i][2], a[i + 2][2] = a[i][0];  
        } maxh = 0; for (i = 0; i < n;  
        ++i) if (Maxh < DAG (i)) maxh = Dp[i];  
    printf ("Case%d:maximum height =%d\n", ++cas, Maxh);  
return 0; }

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

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