UVa 103 Stacking Boxes 解题源码

Problem descriptioin: http://uva.onlinejudge.org/external/1/103.html

采用动态规划记忆化搜索（隐式图）实现

/*
 * 103.cc
 *
 *  Created on: Dec 27, 2012
 *      Author: guixl
 */

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int MAX = 50;
int n, d;
int v[MAX][10];
int graph[MAX][MAX];
int visit[MAX];
int length[MAX];
int pre[MAX];

int Search(int node) {
	if(visit[node])
		return length[node];
	int value = 1;
	for (int i=0; i<n; i++) {
		if (graph[node][i]) {
			int tmp = Search(i) + 1;
			if (tmp > value) {
				value = tmp;
				pre[node] = i;
			}
		}
	}
	length[node] = value;
	visit[node] = 1;
	return value;
}

int main(int argc, char** argv) {
	while (scanf("%d%d", &n, &d) != EOF) {
		for (int i=0; i<n; i++)
			for (int j=0; j<d; j++) {
				scanf("%d", &v[i][j]);
			}
		for (int i=0; i<n; i++) {
			sort(v[i], v[i] + d);
		}
		memset(graph, 0, sizeof(graph));
		for (int i=0; i<n; i++)
			for (int j=0; j<n; j++)
				if (i != j) {
					bool nested = true;
					for (int k=0; k<d; k++)
						if (v[i][k] >= v[j][k]) {
							nested = false;
							break;
						}
					if (nested)
						graph[i][j] = 1;
				}
		//init result store arrays
		for (int i=0; i<n; i++)
			length[i] = 1;
		memset(visit, 0, sizeof(visit));

		for (int i=0; i<n; i++) {
			Search(i);
		}

		int value = 1;
		int start = -1;
		for (int i=0; i<n; i++) {
			if (length[i] > value) {
				value = length[i];
				start = i;
			}
		}
		printf("%d\n", value);
		for (int i=0; i<value; i++) {
			printf("%d ", start+1);
			start = pre[start];
		}
		printf("\n");
	}
	return 0;
}