东华大学2020年程序设计竞赛（同步赛）C.City Supplies

题目链接

题目描述

YZ is the king of the kingdom. There are n cities in his kingdom. To celebrate the 50th anniversary of the founding of his country, YZ decided to distribute supplies as a reward to all the cities.
We all know that the further the city is from the capital (always at 1), the more it will cost to transport. We define K as the shortest distance between a city and the capital, and ignore the difference in distance between different cities(all is 1 unit). Cost from capital to the city is $2^K$2K .
Now YZ gives you the map of his kingdom, and asks you if you can calculate the total cost.
(We guarantee that it is a connected graph.)

输入描述:

The first line contains two integers n and m $(1 \le n \le 10^6,n-1 \le m \le min(2*10^6,(n-1)*n/2))$(1≤n≤106,n−1≤m≤min(2∗106,(n−1)∗n/2)), which means there are n cities and m roads.
The next m lines contains two integers u and v, denoting there is a road connecting city u and city v.

输出描述:

Print the only line containing a single integer. It should be equal to the total cost mod 1e9+7.

示例1

输入

3 2
1 2
2 3

输出

6

示例2

输入

7 6
1 2
1 3
1 4
3 5
3 6
4 7

输出

18

这题不难，用 BFS 求出 1 到每个点的距离，然后跑一遍 O(N) 计算答案即可，数据量大要用快读，AC代码如下:

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N=1e6+5;
const ll mod=1e9+7;
vector<int>g[N];
ll d[N]={0},vis[N]={0};
ll power(ll a,ll b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
void bfs(int u){
    queue<int>q;
    q.push(u);
    while(!q.empty()){
        int a=q.front();
        q.pop();
        for(int b:g[a]){
            if(!vis[b]) {
                d[b]=d[a]+1;
                vis[b]=1;
                q.push(b);
            }
        }
    }
}

inline int read() {
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

 int main(){
     int n,m,x,y;
     n=read();
     m=read();
     while(m--){
        x=read(),y=read();
        g[x].push_back(y);
        g[y].push_back(x);
     }
     bfs(1);
     ll ans=0;
     for(int i=2;i<=n;i++) ans=(ans+power(2,d[i]))%mod;
     cout<<ans;
}

标签：city,le,City,int,ll,Supplies,东华大学,cities,mod
来源： https://blog.csdn.net/qq_43765333/article/details/106607308