Fantasia
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1654 Accepted Submission(s): 429
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
/*
hdu 5739 割点
problem:
给你一个无向图,G[i]为删除i点时,无向图的价值. 求 sum(i*G[i])%mod
如果当前是连通的,那么连通分量的价值为所有点权值的积(任意两个节点连通)
否则为拆分后的各个连通分量的价值的和
solve:
所以需要判断当前点是否是割点.
如果不是割点,只需要减去这个点的权值即可. 如果是割点,要减去这个连通分量的价值再加上拆散后的各个连通分量的值
最开始题意理解错了- -,而且模板有点问题,一直wa.
hhh-2016-08-27 19:47:17
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <math.h>
#include <queue>
#include <map>
#define lson i<<1
#define rson i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define scanfi(a) scanf("%d",&a)
#define scanfl(a) scanf("%I64d",&a)
#define key_val ch[ch[root][1]][0]
#define inf 1e9
using namespace std;
const ll mod = 1e9+7;
const int maxn = 100005;
struct Edge
{
bool cut ;
int v,next,w;
} edge[maxn*5];
int head[maxn],tot;
int low[maxn],dfn[maxn],Stack[maxn],index,top;
bool Instack[maxn],cut[maxn];
int bridge;
ll val[maxn],mul[maxn],ans[maxn],fans[maxn],tval[maxn];
void add_edge(int u,int v)
{
edge[tot].v = v,edge[tot].next = head[u],head[u] = tot++;
}
ll tans =1 ;
ll pow_mod(ll a,ll n)
{
ll cnt =1 ;
while(n)
{
if(n & 1) cnt = cnt*a%mod;
a = a*a%mod;
n >>= 1;
}
return cnt ;
}
int now;
vector<int> vec[maxn];
int from[maxn];
void Tarjan(int u,int ance,int pre)
{
int v;
vec[now].push_back(u);
from[u] = now;
low[u] = dfn[u] = ++index;
Stack[top++] = u;
Instack[u] = true;
tans = tans*val[u] % mod;
int son = 0;
for(int i= head[u]; i!= -1; i = edge[i].next)
{
v = edge[i].v;
if(v == pre){
continue;
}
if(!dfn[v])
{
son ++ ;
ll tp = tans;
Tarjan(v,ance,u);
low[u] = min(low[u],low[v]);
if(u != pre && low[v] >= dfn[u])
{
cut[u] = true;
ll ta = tans * pow_mod(tp,mod-2)%mod;
// cout <<"node:" << u <<" ta:" <<ta <<endl;
ans[u] = (ans[u] + ta)%mod;
fans[u] = (fans[u] * ta) % mod;
}
}
else if(low[u] > dfn[v])
low[u] = dfn[v];
}
if(u == ance && son > 1)
cut[u] = true;
Instack[u] = false;
top --;
}
void init(int n)
{
for(int i = 0; i <= n+1; i++)
{
head[i] = -1;
ans[i] = 0;
fans[i] = 1;
Instack[i]=cut[i]= 0;
dfn[i] = 0;
vec[i].clear();
}
tot=top=index=0;
}
int main()
{
// freopen("in.txt","r",stdin);
int T,n,m,u,v;
scanfi(T);
while(T--)
{
scanfi(n),scanfi(m);
init(n);
for(int i =1; i <= n; i++)
{
scanfl(val[i]);
fans[i] = 1;
}
for(int i = 0; i < m; i++)
{
scanfi(u),scanfi(v);
add_edge(u,v);
add_edge(v,u);
}
now = 1;
ll ob = 0;
for(int i = 1; i <= n; i++)
{
if(!dfn[i])
{
tans= 1;
Tarjan(i,i,-1);
tval[now] = tans;
ll amul = tans;
ob = (ob+tans) %mod;
// cout << "all:" <<tans<<endl;
for(int j = 0 ; j < vec[now].size(); j ++)
{
int to = vec[now][j];
// cout << to <<" " << fans[to] << endl;
if(to == i) continue;
ans[to] = (ans[to] + amul*pow_mod(fans[to]*val[to]%mod,mod-2)%mod);
if(ans[to] > mod) ans[to] -= mod;
}
now ++;
}
}
ll out = 0;
ll tm;
for(int i = 1; i <= n; i++)
{
// cout << fans[i] <<" " << ans[i] <<" " <<cut[i] << endl;
int tf = from[i];
if(cut[i])
{
tm = (ob - tval[tf] + ans[i] + mod)%mod;
}
else
{
if(vec[from[i]].size() > 1)
tm = (ob - tval[tf] + tval[tf]*pow_mod(val[i],mod-2)%mod + mod) % mod;
else
tm = (ob - tval[tf] + mod) % mod;
}
out = (out + i * tm % mod) % mod;
}
printf("%I64d
",out);
}
return 0;
}
/*
3
4 3
1 2 3 4
1 2
2 3
1 3
4 2
100000000 131231232 312354435 432134234
1 2
3 4
66
315142079
*/