51nod 1610 路径计数(dp、容斥)

题意:

$给定一个N\le 100点,M\le 5\times 10^4边的有向无环图$
$一条路径的值:=路径上所有边权的最大公约数$
$Q\le 500次修改操作,每次修改一条边的边权\le 100$
$每次修改后输出有向无环图上路径的值为1的路径数量,答案模10^9+7$

分析:

$修改1条边所能影响的是他的约数那些边,看数据发现每次询问最多只兹磁O(n^2)$
$暴力做是不行的,考虑对倍数容斥一下$
$f[div][i][j]:=值为div倍数的i\to j的方法数$
$根据拓扑序dp一下就得到dp[div][i]:=以i结尾的值为div倍数的方法数$
$g[i]:=值为i的倍数的方法数,h[i]:=值为i的方法数,这个容斥一下就好了$
$对于每次修改显然只影响约数那些,拿出来暴力重新dp,再容斥算答案就好了$
$本题约数级别是sqrt(100)也就是10,所以复杂度是O(100\times n^2+q\times 10\times 100^2)$

代码:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
//
// Created by TaoSama on 2016-08-29
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <stdio.h>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 100 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 5e4 + 10;
typedef long long LL;
int n, m;
LL f[N][N][N], g[N], h[N];
//x倍数的i->j的路径数, x的倍数的路径数, x的路径数
int u[M], v[M], c[M];
vector<int> G[N], topo;
vector<int> divisors[N];
void gao() {
for(int i = 1; i < N; ++i)
for(int j = i; j < N; j += i)
divisors[j].push_back(i);
}
#define rep(i, a, b) for(int i = a; i <= b; ++i)
LL calc(int div) {
LL sum = 0;
vector<LL> dp(n + 1, 0);
for(int i = 0; i < topo.size(); ++i) {
int u = topo[i];
sum = (sum + dp[u]) % MOD;
for(int j = i + 1; j < topo.size(); ++j) {
int v = topo[j];
dp[v] += f[div][u][v] * (1 + dp[u]);
dp[v] %= MOD;
}
}
return sum;
}
LL solve() {
for(int i = 100; i; --i) {
h[i] = g[i];
for(int j = i + i; j <= 100; j += i) h[i] -= h[j];
h[i] = (h[i] % MOD + MOD) % MOD;
}
return h[1];
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
gao();
scanf("%d%d", &n, &m);
vector<int> in(n + 1, 0);
for(int i = 1; i <= m; ++i) {
scanf("%d%d%d", u + i, v + i, c + i);
G[u[i]].push_back(v[i]); ++in[v[i]];
for(int div : divisors[c[i]]) ++f[div][u[i]][v[i]];
}
for(int i = 1; i <= n; ++i) if(!in[i]) topo.push_back(i);
for(int i = 0; i < topo.size(); ++i) {
int u = topo[i];
for(int v : G[u]) if(--in[v] == 0) topo.push_back(v);
}
for(int i = 1; i <= 100; ++i) g[i] = calc(i);
printf("%lld\n", solve());
int q; scanf("%d", &q);
while(q--) {
int x, y; scanf("%d%d", &x, &y);
vector<int> affected;
for(int div : divisors[c[x]]) {
--f[div][u[x]][v[x]];
affected.push_back(div);
}
c[x] = y;
for(int div : divisors[c[x]]) {
++f[div][u[x]][v[x]];
affected.push_back(div);
}
sort(affected.begin(), affected.end());
affected.resize(unique(affected.begin(), affected.end()) - affected.begin());
for(int div : affected) g[div] = calc(div);
printf("%lld\n", solve());
}
return 0;
}


1. 除非注明,本博文即为原创,转载请注明链接地址
2. 本博文只代表博主当时的观点或结论,请不要恶意攻击
3. 如果本文帮到了您,不妨点一下 下面分享到 按钮,让更多的人看到