DP (DAG) UVA 437 the Tower of Babylon

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Test instructions: Give some of the length and width of the bricks, bricks can be stacked on the other side of the long width is smaller than the length of the following head, ask how high can be stacked up

Analysis: Set a brick of the length of the width of x, Y, Z, then want to be more than X, Z, y and y, X, Z of the bricks, if I can be stacked on J, then g[i][j] = True, converted to dag problem, Dp[i] indicates that block I stack at the highest height of the upper

Harvesting: Transforming into a classic model

Code:

/************************************************* author:running_time* Created time:2015-8-28 18:00:01* File Na Me:UVA_437.cpp ************************************************/#include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string > #include <vector> #include <queue> #include <deque> #include <stack> #include <list># Include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime>using namespace std; #define Lson L, Mid, RT << 1#define Rson mid + 1, R, RT << 1 | 1typedef long ll;const int N = 1e2 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;struct block{int x, Y, Z;}    B[n];bool g[n][n];int dp[n];int n;int DFS (int u) {if (Dp[u]! =-1) return dp[u]; Dp[u] = b[u].z;for (int i=1; i<=n; ++i) {if (G[u][i]) {Dp[u] = max (Dp[u], DFS (i) + b[u].z);}} return dp[u];} BOOL Check (int I, Int j) {if (b[i].x < b[j].x && B[i].y < B[J].Y) return True;if (b[i].x < b[j].y && B[i].y < b[j].x) return True;return false;} int main (void) {int cas = 0;while (scanf ("%d", &n) = = 1) {if (n = = 0) break;for (int i=1; i<=n; ++i) {scanf (" %d%d%d ", &b[i].x, &b[i].y, &b[i].z); b[n+i].x = b[i].x, b[n+i].y = b[i].z, b[n+i].z = b[i].y;b[2*n+i].x = B[i]. Y, b[2*n+i].y = b[i].z, b[2*n+i].z = b[i].x;} Memset (g, false, sizeof (g)), n *= 3;for (int i=1; i<=n; ++i) {for (int j=i+1; j<=n; ++j) {if (check (i, j)) G[i][j] = True;if (Check (J, i)) g[j][i] = true;}} Memset (DP,-1, sizeof (DP)), int ans = 0;for (int i=1; i<=n; ++i) {ans = max (ans, DFS (i));}    printf ("Case%d:maximum height =%d\n", ++cas, ans);} return 0;}

　　

DP (DAG) UVA 437 the Tower of Babylon