【HDOJ6578】Blank（DP）

题意：一个长为n的序列，每个位置上的值是0,1,2,3中的一个，有m个限制条件，限制位置[l[i],r[i]]中不同的数值有x[i]个，问方案数MOD 998244353

n<=100,m<=100

思路：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,int>P;
 #define N  110
 #define M  151000
 #define fi first
 #define se second
 #define MP make_pair
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const int MOD=,inv2=(MOD+)/;
       double eps=1e-;
       ll INF=1e18;
       ll inf=5e13;
       int dx[]={-,,,};
       int dy[]={,,-,};

 int dp[][N][N][N];
 VII c[N];

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void calc(int &a,int b)
 {
     a+=b;
     if(a>=MOD) a-=MOD;
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     int cas=read();
     while(cas--)
     {
         int n=read(),m=read();
         rep(i,,n) c[i].clear();
         rep(i,,m)
         {
             int x=read(),y=read(),z=read();
             //printf("x=%d y=%d z=%d\n",x,y,z);
             c[y].push_back(MP(x,z));
         }
         rep(i,,n)
          rep(j,,n)
           rep(k,,n) dp[][i][j][k]=;
         dp[][][][]=;
         int v=;
         rep(i,,n)
         {
             v^=;
             rep(j,,i)
              rep(k,,j)
               rep(t,,k) dp[v][j][k][t]=;
             rep(j,,i-)
              rep(k,,j)
               rep(t,,k)
               {
                     calc(dp[v][j][k][t],dp[-v][j][k][t]);
                     calc(dp[v][i-][k][t],dp[-v][j][k][t]);
                     calc(dp[v][i-][j][t],dp[-v][j][k][t]);
                     calc(dp[v][i-][j][k],dp[-v][j][k][t]);
               }
             rep(j,,i-)
              rep(k,,j)
               rep(t,,k)
               {
                   for(int x=;x<c[i].size();x++)
                   {
                       PII tmp=c[i][x];
                       if((j>=tmp.fi)+(k>=tmp.fi)+(t>=tmp.fi)+!=tmp.se) dp[v][j][k][t]=;
                   }
               }
         }
         int ans=;
         rep(i,,n-)
          rep(j,,i)
           rep(k,,j) calc(ans,dp[v][i][j][k]);
         printf("%d\n",ans);

     }
     return ;
 }