Description
给你一个长度为 N 的序列和 M 组约束条件 , 每组条件形如L i , R i , X i , 表示序列上的 [L i , R i ] 中恰好有 X i 种颜色 , 现在要你用三种颜
色给这个序列染色 , 求满足所有约束的方案数 .
Solution
用dpt,i,j,kdpt,i,j,k表示到第tt位,上一次出现红色/绿色/蓝色的位置为,转移非常显然。我们发现t=max(i,j,k)t=max(i,j,k),可以省掉一维。
注意统计答案时要判断是否合法(由于这个一直没有过样例)。
Code
/************************************************
* Au: Hany01
* Date: Aug 23rd, 2018
* Prob: ARC074 E
* Email: hany01dxx@gmail.com & hany01@foxmail.com
* Inst: Yali High School
************************************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia
template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }
inline int read() {
static int _, __; static char c_;
for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}
const int maxn = 305, MOD = 1e9 + 7;
int n, m, l, r, x, dp[maxn][maxn][maxn], Ans;
vector<PII> lmt[maxn];
inline void ad(int& x, int y) { if ((x += y) >= MOD) x -= MOD; }
int main()
{
#ifdef hany01
freopen("arc074e.in", "r", stdin);
freopen("arc074e.out", "w", stdout);
#endif
n = read(), m = read();
For(i, 1, m) l = read(), r = read(), x = read(), lmt[r].pb(mp(l, x));
dp[0][0][0] = 1;
For(i, 0, n) For(j, 0, i) For(k, 0, i) if (dp[i][j][k]) {
rep(p, SZ(lmt[i])) {
int sm = 1, l = lmt[i][p].x;
if (j >= l) ++ sm;
if (k >= l) ++ sm;
if (sm != lmt[i][p].y) { dp[i][j][k] = 0; break; }
}
if (!dp[i][j][k]) continue;
if (i == n) ad(Ans, dp[i][j][k]);
else {
ad(dp[i + 1][j][k], dp[i][j][k]);
ad(dp[i + 1][i][j], dp[i][j][k]);
ad(dp[i + 1][i][k], dp[i][j][k]);
}
}
printf("%d\n", Ans);
return 0;
}