[2019HDU first correction Multi] [HDU 6578] [A. Blank]

Topic links: http://acm.hdu.edu.cn/showproblem.php?pid=6578

Title effect: length \ (n-\) are filled array requires \ (\ {0,1,2,3 \} \) according to any one of four numbers, there are \ (m \) limitation conditions: Interval digital several species \ ([l, r] \) appears exactly \ (x \), find the number of programs

Interpretations: f [i] [j] [k] [cur] represents the number of the last four positions occurring after sorting number \ (i, j, k, cur \) program, can transfer force, wherein a final dimensional array to scroll to save space

　　　For vector constraints can survive, the right end point of each cycle is determined to be a restriction for the current point

　　　Although four sets to \ (for \), but due to the sequential limitation, the execution times of approximately \ (\ frac {n ^ 4} {24} \), coupled with a simple operation of this problem, a small constant, substantially TLE situation does not appear

 

　　　But still have to Tucao a topic and people have time to set up the game good ... Sia open 3s should not have had it orz holiday algorithm

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define N 101
 4 #define mp make_pair
 5 #define MOD 998244353
 6 int T,n,m,l,r,x,f[N][N][N][2],ans;
 7 vector<pair<int,int>>d[N];
 8 void init()
 9 {
10     ans=0;
11     scanf("%d%d",&n,&m);
12     for(int i=1;i<=n;i++)
13       {
14       d[i].clear();
15       d[i].push_back(mp(i,1));
16       }
17     for(int i=1;i<=m;i++)
18       {
19       scanf("%d%d%d",&l,&r,&x);
20       d[r].push_back(mp(l,x));
21       }
22     memset(f,0,sizeof(f));
23     f[0][0][0][0]=1;
24     for(int cur=1;cur<=n;cur++)
25       {
26       int o=cur&1;
27       for(int i=0;i<=cur;i++)
28         for(int j=i;j<=cur;j++)
29           for(int k=j;k<=cur;k++)
30             f[i][j][k][o]=0;
31       for(int i=0;i<=cur;i++)
32         for(int j=i;j<=cur;j++)
33           for(int k=j;k<=cur;k++)
34             {
35             (f[j][k][cur-1][o]+=f[i][j][k][o^1])%=MOD;
36             (f[i][k][cur-1][o]+=f[i][j][k][o^1])%=MOD;
37             (f[i][j][cur-1][o]+=f[i][j][k][o^1])%=MOD;
38             (f[i][j][k][o]+=f[i][j][k][o^1])%=MOD;
39             }
40       for(int i=0;i<=cur;i++)
41         for(int j=i;j<=cur;j++)
42           for(int k=j;k<=cur;k++)
43             for(auto pi:d[cur])
44               {
45               l=pi.first,r=cur,x=pi.second;
46               if((i>=l)+(j>=l)+(k>=l)+(cur>=l)!=x)
47                 f[i][j][k][o]=0;
48               }
49       }
50     for(int i=0;i<=n;i++)
51       for(int j=i;j<=n;j++)
52         for(int k=j;k<=n;k++)
53           (ans+=f[i][j][k][n&1])%=MOD;
54     printf("%d\n",ans);
55 }
56 int main()
57 {
58     scanf("%d",&T);
59     while(T--)init();
60 }

View Code

Code to initialize the vector is not really necessary, additional push_back, writing for insurance when you go to add (although almost led TLE)

 

It turned out to be my blog in the first line of pure DP problem ...?