东华大学2020年程序设计竞赛（同步赛） C题 City Supplies 最短路迪杰斯特拉+堆优化

链接：https://ac.nowcoder.com/acm/contest/5891/C
来源：牛客网

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 2K2^K2K.
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≤n≤106,n−1≤m≤min(2∗106,(n−1)∗n/2))(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
复制

3 2
1 2
2 3

输出
复制

示例2
输入
复制

输出
复制

题意 : 给定一张无向图,源点为 $1$ , 两点之间的边权为 $2^{K}$ , 源点到其他点求边权和mod1e9+7
如图
注意点 :

要预处理出所有 $2^{k}$ (一开始直接暴力快速幂 $T L E$ 了)

#define debug
#ifdef debug
#include <time.h>
#include </home/majiao/mb.h>
#endif

#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <math.h>

#define MAXN ((int)1e6+7)
#define ll long long 
#define INF (0x7f7f7f7f)
#define fori(lef, rig) for(int i=lef; i<=rig; i++)
#define forj(lef, rig) for(int j=lef; j<=rig; j++)
#define fork(lef, rig) for(int k=lef; k<=rig; k++)
#define QAQ (0)

using namespace std;

#define show(x...) \ do { \ cout << "\033[31;1m " << #x << " -> "; \ err(x); \ } while (0)

void err() { cout << "\033[39;0m" << endl; }
template<typename T, typename... A>
void err(T a, A... x) { cout << a << ' '; err(x...); }

namespace FastIO{

	char print_f[105];
	void read() {}
	void print() { putchar('\n'); }

	template <typename T, typename... T2>
	   inline void read(T &x, T2 &... oth) {
		   x = 0;
		   char ch = getchar();
		   ll f = 1;
		   while (!isdigit(ch)) {
			   if (ch == '-') f *= -1; 
			   ch = getchar();
		   }
		   while (isdigit(ch)) {
			   x = x * 10 + ch - 48;
			   ch = getchar();
		   }
		   x *= f;
		   read(oth...);
	   }
	template <typename T, typename... T2>
	   inline void print(T x, T2... oth) {
		   ll p3=-1;
		   if(x<0) putchar('-'), x=-x;
		   do{
				print_f[++p3] = x%10 + 48;
		   } while(x/=10);
		   while(p3>=0) putchar(print_f[p3--]);
		   putchar(' ');
		   print(oth...);
	   }
} // namespace FastIO
using FastIO::print;
using FastIO::read;

int n, m, Q, K;

struct Edge {
	int to, val;
} ;

typedef pair<int, int> pii;
vector<Edge> G[MAXN];

int d[MAXN];
void dij() {
	int s = 1;
	memset(d, INF, sizeof(d));
	priority_queue<pii> q;
	q.push({-d[s], s}); d[s] = 0;
	while(!q.empty()) {
		auto now = q.top(); q.pop();
		int u = now.second;
		for(auto ed : G[u]) {
			int v = ed.to, td = ed.val;
			if(d[v] > d[u]+td) {
				d[v] = d[u] + td;
				q.push({-d[v], v});
			}
		}
	}
}
ll len[MAXN] = { 0 };

signed main() {
#ifdef debug
	freopen("test", "r", stdin);
	clock_t stime = clock();
#endif
	scanf("%d %d ", &n, &m);
	while(m--) {
		int u, v;
		scanf("%d %d ", &u, &v);
		G[u].push_back({v, 1}), G[v].push_back({u, 1});
	}
	dij();
	int tmax = 0;
	for(int i=1; i<=n; i++) tmax = max(tmax, d[i]);
	len[0] = 1;
	int mod = 1e9+7;
	for(int i=1; i<=tmax; i++) {
		len[i] = len[i-1] << 1;
		if(len[i] >= mod) //不优化这里不给过 ???
			len[i] %= mod;
	}
	ll ans = 0;
	for(int i=2; i<=n; i++) {
		ans += len[d[i]]; 
		if(ans >= mod) ans %= mod; ////不优化这里不给过 ???
	}
	printf("%lld\n", ans);



#ifdef debug
	clock_t etime = clock();
	printf("rum time: %lf 秒\n",(double) (etime-stime)/CLOCKS_PER_SEC);
#endif 
	return 0;
}