uva 103 Stacking Boxes（DAG）

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题目连接：103 - Stacking Boxes

题目大意：有n个w维立体，输出立体互相嵌套的层数的最大值，并输出嵌套方式，可嵌套的要求是外层立体的w条边可以分别对应大于内层立体。

解题思路：可以将每个立体的w边安照从小到大排序，比较的时候逐一比较就可以判断出可否嵌套，然后就是纯DAG问题了。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 50;int n, w, dp[N], vis[N], seq[N][N];void Init() {    memset(dp, 0, sizeof(dp));    memset(vis, 0, sizeof(vis));    memset(seq, 0, sizeof(seq));}void read() {    for (int i = 0; i < n; i++) {for (int j = 0; j < w; j++)    scanf("%d", &seq[i][j]);sort(seq[i], seq[i] + w);    }}bool judge(int a, int b) {    for (int i = 0; i < w; i++)if(seq[a][i] <= seq[b][i])  return false;    return true;}int find(int d) {    if (dp[d])return dp[d];    for (int i = 0; i < n; i++) {if (judge(i, d)) {    int a = find(i);    if (a > dp[d]) {dp[d] = a;vis[d] = i;    }}    }    return ++ dp[d];}void solve() {    int Max = 0, id, a;    for (int i = 0; i < n; i++) {a = find(i);if (a > Max) {    Max = a;    id = i;}    }    printf("%d\n", Max);    for (int i = 1; i < Max; i++) {printf("%d ", id + 1);id = vis[id];    }    printf("%d\n", id + 1);}int main() {    while (scanf("%d%d", &n, &w) == 2) {Init();read();solve();    }    return 0;}