hdu5739Fantasia(多校第二场1006) 割点+逆元
Fantasia
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Problem Description
Professor Zhang has an undirected graph G with n vertices and m edges. Each vertex is attached with a weight wi. Let Gi be the graph after deleting the i-th vertex from graph G. Professor Zhang wants to find the weight of G1,G2,...,Gn.
The weight of a graph G is defined as follows:
1. If G is connected, then the weight of G is the product of the weight of each vertex in G.
2. Otherwise, the weight of G is the sum of the weight of all the connected components of G.
A connected component of an undirected graph G is a subgraph in which any two vertices are connected to each other by paths, and which is connected to no additional vertices in G.
The weight of a graph G is defined as follows:
1. If G is connected, then the weight of G is the product of the weight of each vertex in G.
2. Otherwise, the weight of G is the sum of the weight of all the connected components of G.
A connected component of an undirected graph G is a subgraph in which any two vertices are connected to each other by paths, and which is connected to no additional vertices in G.
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains two integers n and m (2≤n≤105,1≤m≤2×105) -- the number of vertices and the number of edges.
The second line contains n integers w1,w2,...,wn (1≤wi≤109), denoting the weight of each vertex.
In the next m lines, each contains two integers xi and yi (1≤xi,yi≤n,xi≠yi), denoting an undirected edge.
There are at most 1000 test cases and ∑n,∑m≤1.5×106.
The first line contains two integers n and m (2≤n≤105,1≤m≤2×105) -- the number of vertices and the number of edges.
The second line contains n integers w1,w2,...,wn (1≤wi≤109), denoting the weight of each vertex.
In the next m lines, each contains two integers xi and yi (1≤xi,yi≤n,xi≠yi), denoting an undirected edge.
There are at most 1000 test cases and ∑n,∑m≤1.5×106.
Output
For each test case, output an integer S=(∑i=1ni⋅zi) mod (109+7), where zi is the weight of Gi.
Sample Input
1
3 2
1 2 3
1 2
2 3
Sample Output
20
题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=5739
题目描述：
有n个节点，每个节点有个值，然后m条边构成可能不止一张张图，每张图的价值是每个节点的值的乘积，然后总的价值就是所有图的价值加起来。
现在要分别删除每个点，G1就是代表的删除编号为1的节点，所有图加起来的价值。 然后问你1*G[2] + 2*G[2] + ……+n*G[n]. 最后的值膜一个1e9+7。
题解：
这个题很显然就是求割点，如果不是割点，就直接删除这个点就好了，如果是割点就复杂一点，就需要将该点的子树中最多能够访问到该点的子树的值给处理出来。然后把子树分开，另外处理就好了，还是看代码分析吧。道理是这么说，但是中途写错了好多东西TAT，对着数据改到现在...唉....
代码：
#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>
#include<queue>
#include<set>
#include<string>
#include<map>
#define inf 9223372036854775807
#define INF 9e7+5
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const db eps = 1e-9;
ll va[maxn], w[maxn], Sum, ans[maxn];
int pre[maxn], dfs_tim, tot, n, m, low[maxn], t, vep[maxn];
bool vis[maxn];
vector<int> G[maxn];
void init() {
memset(vis, false, sizeof(vis));
memset(pre, 0, sizeof(pre));
Sum = tot = dfs_tim = 0;
for (int i = 1; i <= n; i++) G[i].clear();
}
//快速幂，求逆元用
ll pow_mod(ll a, ll b, ll p) {
ll ret = 1;
while(b) {
if(b & 1) ret = (ret * a) % p;
a = (a * a) % p;
b >>= 1;
}
return ret;
}
//费马小定理求的逆元
ll inv(ll x) {
return pow_mod(x, mod-2, mod);
}
// 先写好，懒得每次模
void add(ll &x, ll y) {
x = x + y;
x = (x + mod) % mod;
}
// 主要是把每张图的价值处理出来
void Find(int x) {
va[x] = w[x];
for (int i = 0; i < G[x].size(); i++) {
int u = G[x][i];
if (vis[u]) continue;
vis[u] = true; Find(u);
va[x] = va[x] * va[u] % mod;
}
}
ll dfs(int x, int fa, int root) { //当前节点，父节点和根节点
low[x] = pre[x] = ++dfs_tim; //pre数组记录访问的时间
ans[x] = inv(w[x]); //删除此时访问的节点
int cld = 0; ll sum = 0, res = w[x], pro = 1;
for (int i = 0; i < G[x].size(); i++) {
int u = G[x][i];
if (!pre[u]) {
cld++;
ll tmp = dfs(u, x, root); //tmp返回的是对于u这颗子树的价值
low[x] = min(low[x], low[u]); //更新x节点所能访问的最早的祖先
if (low[u] >= pre[x]) { //如果u这颗子树所能访问的是x，那么说明x节点被删除，u这颗子树会被分开
add(sum, tmp); //sum表示的是x节点被删除后，x会被分开的子树的价值之和
ans[x] = ans[x] * inv(tmp) % mod; //和上面删除节点一样，表示将这颗子树删除
}
res = res * tmp % mod; //求子树的价值
}
else if (u != fa) low[x] = min(low[x], pre[u]); //对于访问比当前节点早的节点，更新能访问的最早节点
} //tt表示的是除了这幅图，其它图的价值之和
ll tt = (Sum - va[root] + mod) % mod; //va[roor]*ans[x]中ans[x]已经是逆元了,所以这句话
ans[x] = va[root] * ans[x] % mod; //表示的是将x节点和会分开的子树 删除后该图的值
if (fa == -1 && ans[x] == 1) ans[x] = 0; //对于一张图，如果他的子节点全部被删除了，我们
//求到的ans[x]是1，但事实上应 该是0，所以
//需要特判一下，比如这样一张图 1 - 2， 1 - 3.
add(ans[x], tt); add(ans[x], sum); //将其他图和删除的子树加起来
if (fa == -1) {
if (cld == 1) { //对于最开始的祖先，如果他只有一个儿
//子，那么他不是割点，学割点应该都学过QAQ
ans[x] = va[root] * inv(w[x]) % mod;
add(ans[x], tt);
}
else if (G[x].size() == 0) {
ans[x] = tt; //如果这是一个孤立点，删除后就直接是其他图的值
}
}
return res;
}
void solve() {
cin >> n >> m;
init();
for (int i = 1; i <= n; i++) scanf("%I64d", &w[i]);
for (int i = 1; i <= m; i++) {
int u, v; scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
for (int i = 1; i <= n; i++) {
if (vis[i]) continue;
vis[i] = true;
vep[++tot] = i; Find(i); //vep数组用来存每次要访问的图的开始节点
add(Sum, va[i]); //所有图的总价值,va[i]就代表了这张图的总价值
}
for (int i = 1; i <= tot; i++) {
dfs(vep[i], -1, vep[i]); //-1位置代表的父节点，对于最开始的点的父亲设为-1
}
ll pri = 0;
for (ll i = 1; i <= n; i++) {
add(pri, i*ans[i]%mod); //求出最后的值
}
cout << pri << endl;
}
int main() {
//cin.sync_with_stdio(false);
// freopen("tt.txt", "r", stdin);
//freopen("hh.txt", "w", stdout);
cin >> t;
while (t--)
solve();
return 0;
}
