HDU 2019 Multi-University Training Contest 1 杭电2019暑期多校集训第一场 1001 Blank (6578)
Problem Description
There are N blanks arranged in a row. The blanks are numbered 1,2,…,N from left to right.
Tom is filling each blank with one number in {0,1,2,3}. According to his thought, the following M conditions must all be satisfied. The i-th condition is:
There are exactly xi different numbers among blanks ∈[li,ri].
In how many ways can the blanks be filled to satisfy all the conditions? Find the answer modulo 998244353.
Input
The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.
In each test case, there are two integers n(1≤n≤100) and m(0≤m≤100) in the first line, denoting the number of blanks and the number of conditions.
For the following m lines, each line contains three integers l,r and x, denoting a condition(1≤l≤r≤n, 1≤x≤4).
Output
For each testcase, output a single line containing an integer, denoting the number of ways to paint the blanks satisfying all the conditions modulo 998244353.
Sample Input
2
1 0
4 1
1 3 3
Sample Output
4
96
题意
有一个长度为n的串，这个串仅由0，1，2，3构成，给定m个条件，条件形式为在l到r区间内有x种数字（1<=x<=4）。问有多少种这样的串，答案模998244353。
思路
DP。定义一个vector记录m个条件。定义一个思维的dp数组，前三维记录每种数字最后出现的位置，最后一种数字出现的位置即为当前字符串长度。第四维为滚动数组，降低空间复杂度。dp时分四种情况进行dp，下一位数字和上一位相同，即直接将dp[i][j][k][last]继承，或者与前三维某个数字相同，将其他数字重新放入这三维进行dp即可。dp完后需判断每种情况是否满足这m个条件，即判断(i>=l)+(j>=l)+(k>=l)+(cur>=l)==x是否成立，若不成立，则将dp值直接刷新为0。最后对dp数组求和即可。
坑点
字符串长度每增加一位就需要将dp数组第四维为now的数组清空。否则就会继续继承上上次长度dp时的dp值。
最后求和得出答案时只需要对第四维为n&1的数组求和。
别忘了模998244353。
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD=998244353;
const int N=105;
ll dp[N][N][N][2];
void init()
{
memset(dp,0,sizeof(dp));
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m;
scanf("%d%d",&n,&m);
vector<pair<int,int> > G[N];
init();
for(int i=0;i<m;i++)
{
int l,r,x;
scanf("%d%d%d",&l,&r,&x);
G[r].push_back(make_pair(l,x));
}
dp[0][0][0][0]=1;
for(int cur=1;cur<=n;cur++)
{
int now=cur&1;
int last=(!now);
for(int i=0;i<=cur;i++)
{
for(int j=i;j<=cur;j++)
{
for(int k=j;k<=cur;k++)
{
dp[i][j][k][now]=0;
}
}
}
for(int i=0;i<=cur;i++)
{
for(int j=i;j<=cur;j++)
{
for(int k=j;k<=cur;k++)
{
dp[i][j][k][now]=(dp[i][j][k][now]+dp[i][j][k][last])%MOD;
dp[i][j][cur-1][now]=(dp[i][j][cur-1][now]+dp[i][j][k][last])%MOD;
dp[j][k][cur-1][now]=(dp[j][k][cur-1][now]+dp[i][j][k][last])%MOD;
dp[i][k][cur-1][now]=(dp[i][k][cur-1][now]+dp[i][j][k][last])%MOD;
}
}
}
for(int i=0;i<=cur;i++)
{
for(int j=i;j<=cur;j++)
{
for(int k=j;k<=cur;k++)
{
for(int o=0;o<(int)G[cur].size();o++)
{
int l=G[cur][o].first;
int r=cur;
int x=G[cur][o].second;
if((i>=l)+(j>=l)+(k>=l)+(cur>=l)!=x)
{
dp[i][j][k][now]=0;
}
}
}
}
}
}
ll ans=0;
for(int i=0;i<=n;i++)
{
for(int j=i;j<=n;j++)
{
for(int k=j;k<=n;k++)
{
ans=(ans+dp[i][j][k][n&1])%MOD;
}
}
}
printf("%lld\n",ans%MOD);
}
return 0;
}
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skr~
