uva103 - Stacking Boxes(动归，记忆化搜索)

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利用条件建图，然后在图中找到最长路径，

状态：dp[i]表示以i为起点的最长路径长度。

状态转移：dp[i] = {max(dp[j]+1)|g[i][j] = 1}

代码如下：

#include <cstdio>#include <cstdlib>#include <cstring>#define M 15#define N 35int m, n, a[N][M], g[N][N], d[N];int comp(const void *aa, const void *bb){    return *(int*)aa-*(int*)bb;}int judge(int s, int t){    for(int i = 0; i < m; i++)    if(a[s][i]>=a[t][i]) return 0;    return 1;}int dp(int x){    int &ans = d[x];    if(ans) return ans;    ans = 1;    for(int i = 1; i <= n; i++)    {        if(g[x][i]&&ans<dp(i)+1) ans = dp(i)+1;    }    return ans;}void print_ans(int x, int cur){    if(cur==1) {printf("%d\n",x); return; }    else printf("%d ",x);    for(int i = 1; i <= n; i++) if(g[x][i]&&d[x]==d[i]+1) {print_ans(i, cur-1); break;}}int main (){    while(~scanf("%d %d",&n,&m))    {        for(int i = 1; i <= n; i++)        {            for(int j = 0; j < m; j++) scanf("%d",&a[i][j]);            qsort(a+i,m,sizeof(int),comp);        }        for(int i = 1; i <= n; i++)        for(int j = 1; j <= n; j++)        {            if(i!=j&&judge(i,j)) g[i][j] = 1;            else g[i][j] = 0;        }        memset(d,0,sizeof(d));        for(int i = 1; i <= n; i++) dp(i);        int max = 1, maxl = 0;        for(int i = 1; i <= n; i++) if(d[max]<d[i]) max = i;        maxl = d[max];        printf("%d\n",maxl);        print_ans(max,maxl);    }    return 0;}