Time limit 1000 ms
Memory limit 262144 kB
OS Windows
Source 2019 Multi-University Training Contest 1
Impressions
On the train AK dls game, live in Topic Two days later, talked about this problem when he said easily, ah, a simple dp get away, then say what parameters dp, and then on to the next the problem ......
Problem-solving ideas
Throw a few links, digest it
- https://blog.csdn.net/m0_43448982/article/details/96988882
- https://blog.csdn.net/Ratina/article/details/97237438
- https://www.cnblogs.com/zhuyou/p/11273721.html
Probably set \ (dp [x] [y ] [z] [i] \) represent the following meaning: front to back sweep, the sweep of the current \ (I \) when the grid, the last four colors to appear They were (x, y, z, i \) \ number scheme, where \ (X \ leqslant Y \ leqslant Z \ leqslant I \) , in \ (x, y, z, i \) is taken as 0 equal sign, he said there are some colors that has not yet appeared, ie the position of the last occurrence is zero.
The kind of color is not important, because the calculation process has been put into the calculation of combinations of different colors, the color is important. This then stained cells ( \ (i \) number scheme) when is contracted from a grid ( \ (i-1 \) Transfer) scheme over the number.
扩展当前这一个格子的时候，我们可以取之前出现过的4种颜色中的一种。举个例子，假设我们取的是之前最后一次出现在\(x\)的那个颜色，那么会发生这样的转移——\(dp[y][z][i-1][i]+=dp[x][y][z][i-1]\)（想想dp数组四个参数的含义和大小关系，现在这个状态的数量增加了上一个状态的数量那么多种），取的是最后一次出现在\(y\)的那个颜色的话，发生的转移就是——\(dp[x][z][i-1][i]+=dp[x][y][z][i-1]\)，同样的道理还有\(dp[x][y][i-1][i]+=dp[x][y][z][i-1]\)和\(dp[x][y][i-1][i]+=dp[x][y][z][i]\)。对于判断条件，当我们跑完第\(i\)个格子的时候，就检查以\(i\)为右端点的所有条件，对于一个条件——\([l,r]\)之间有\(x\)种颜色，我们可以看那些状态是否满足条件，看每个颜色最后一次出现的位置在区间内还是区间外，然后根据这个统计区间内颜色数量，不满足条件就把它们清零，防止转移到下一个\(i\)。
统计的时候就统计所有\(i=n\)的格子，把数量加起来就好。但是这种方法会爆空间，注意到\(i\)那一维可以滚动，所以把\(i\)那一维滚动起来。另外注意取模 993244853 998244353。
源代码
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
const int mod=998244353;
int T;
int n,m;
long long dp[105][105][105][2];
//x<y<z<i
struct Condition{
int l,x;
};
std::vector<Condition> c[105];
int main()
{
//freopen("test.in","r",stdin);
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0;i<105;i++) c[i].clear();
for(int i=1,l,r,x;i<=m;i++)
{
scanf("%d%d%d",&l,&r,&x);
c[r].push_back({l,x});
}
dp[0][0][0][0]=1;
int cur=1;//i&1
int last=0;//!(i&1)
for(int i=1;i<=n;i++)
{
for(int z=0;z<=i;z++)
{
for(int y=0;y<=z;y++)
{
for(int x=0;x<=y;x++)
{
dp[x][y][z][cur]=0;//清零
}
}
}
for(int z=0;z<i;z++)
{
for(int y=0;y<=z;y++)
{
for(int x=0;x<=y;x++)//转移
{
long long temp=dp[x][y][z][last];
dp[x][y][z][cur]+=temp;
dp[y][z][i-1][cur]+=temp;
dp[x][z][i-1][cur]+=temp;
dp[x][y][i-1][cur]+=temp;
dp[x][y][z][cur]%=mod;
dp[y][z][i-1][cur]%=mod;
dp[x][z][i-1][cur]%=mod;
dp[x][y][i-1][cur]%=mod;
}
}
}
for(unsigned int cc=0;cc<c[i].size();cc++)//判断
{
for(int z=0;z<i;z++)
for(int y=0;y<=z;y++)
for(int x=0;x<=y;x++)
{
int temp=1+(x>=c[i][cc].l?1:0)+(y>=c[i][cc].l?1:0)+(z>=c[i][cc].l?1:0);//统计区间内颜色数量
if(temp!=c[i][cc].x) dp[x][y][z][cur]=0;
}
}
std::swap(cur,last);
}
long long ans=0;
for(int z=0;z<n;z++)//统计答案
{
for(int y=0;y<=z;y++)
{
for(int x=0;x<=y;x++)
{
ans+=dp[x][y][z][last];
ans%=mod;
}
}
}
printf("%lld\n",ans);
}
return 0;
}