CF 938D Buy a Ticket

算法标签: 最短路、图论

题目

题目描述

流行乐队“Flayer”将在n个城市开演唱会这n个城市的人都想去听演唱会每个城市的票价不同于是这些人就想是否能去其他城市听演唱会更便宜（去要路费的）输入格式：第一行包含两个整数n和m 接下来m行每行三个数 u v k 表示u城市到v城市要k元接下来n个数表每个城市的票价

输入格式

The first line contains two integers (n) and (m) (( 2<=n<=2·10^{5} , 1<=m<=2·10^{5} )).

Then (m) lines follow,$ i $-th contains three integers (v_{i}) , (u_{i}) , and (w_{i}) (( 1<=v_{i},u_{i}<=n,v_{i}≠u_{i} , 1<=w_{i}<=10^{12} )) denoting (i) -th train route. There are no multiple train routes connecting the same pair of cities, that is, for each ((v,u))neither extra$ (v,u)$ nor$ (u,v) $present in input.

The next line contains nn integers (a_{1},a_{2},... a_{k} ( 1<=a_{i}<=10^{12} )) — price to attend the concert in (i)-th city.

输出格式

Print (n) integers. (i) -th of them must be equal to the minimum number of coins a person from city $ i$ has to spend to travel to some city $ j$ (or possibly stay in city $ i $), attend a concert there, and return to city (i (if~~ j≠i )).

输入输出样例

输入 #1

输出 #1

6 14 1 25

输入 #2

输出 #2

12 10 12

题解：

最短路+超级源点

为什么在考场上想了那么久 ———— 《打脸》

这种描述很轻松就可以看出来这是一道略微不太正经的最短路，不正经在哪？？？

开完演唱会还需要回到原来的城市——二倍边权
在每个点开演唱会有一定的点权——(Important)

二倍边权这个怎么处理不用说了吧………………

问题就在于如何处理点权？？每个点都跑一下Dij？？

——显然做不到

考虑超级源点，把所有点都与超级源点连接一条边权为该点点权的边，对于超级源点跑最短路。（除了超级源点与每个点之间的边以外的所有边都要存二倍边权）。

这样跑完直接就是结果了。

AC代码

#include <bits/stdc++.h>

#define setI(x) freopen(x".in", "r", stdin);

#define setO(x) freopen(x".out", "w", stdout);

#define setIO(x) setI(x) setO(x)

using namespace std;

typedef long long ll;

char *p1, *p2, buf[100000];

#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )

int rd() {
	int x = 0, f = 1;
	char c = nc();
	while (c < 48) {
		if (c == '-') {
			f = -1;
		}
		c = nc();
	}
	while (c > 47) {
		x = (((x << 2) + x) << 1) + (c ^ 48);
		c = nc();
	}
	return x * f;
}

ll lrd() {
	ll x = 0, f = 1;
	char c = nc();
	while (c < 48) {
		if (c == '-') {
			f = -1;
		}
		c = nc();
	} 
	while (c > 47) {
		x = (((x << 2) + x) << 1) + (c ^ 48);
		c = nc();
	}
	return x * f;
}

const int N = 2e5 + 10;

const int inf = 0x3f3f3f3f;

int tot, head[N], to[N << 2], nxt[N << 2];

ll val[N << 2];

int n, m;

ll num[N];

ll dis[N];

bool vis[N];

void add(int x, int y, ll z) {
	to[ ++ tot] = y;
	val[tot] = z;
	nxt[tot] = head[x];
	head[x] = tot;
}

priority_queue < pair<ll, int> > q;

void dijkstra(int s)
{
	memset(dis, 0x3f, sizeof dis);
	// memset(vis, 0, sizeof vis);
	dis[s] = 0;
	q.push(make_pair(0, s));
	while (!q.empty())
	{
		if (vis[q.top().second])
		{
			q.pop();
			continue ;
		}
		int x = q.top().second;
		q.pop();
		vis[x] = 1;
		for (int i = head[x]; i; i = nxt[i])
		{
			if (dis[to[i]] > dis[x] + val[i])
			{
				dis[to[i]] = dis[x] + val[i];
				q.push(make_pair(-dis[to[i]], to[i]));
			}
		}
	}
}

int main() {

// setIO("movie");
	// scanf("%d%d", &n, &m);
	n = rd(), m =rd();
	for (int i = 1; i <= m; i ++ ) {
		int a = rd(), b = rd();
		ll c = lrd();
		add(a, b, c * 2);
		add(b, a, c * 2);
	}
	for (int i = 1; i <= n; i ++ ) {
		num[i] = lrd();
		add(0, i, num[i]);
	}
	dijkstra(0);
	for (int i = 1; i <= n; i ++ ) {
		printf("%lld ", dis[i]);
	}
	return 0;

}