LuoguP1629 邮递员送信 题解

这里先说个题外话——

正如我在此题讨论区里发的一个帖子所说，这个题目和P1342 请柬是一样的，但是需要开大数组，并将答案开\(long~long\)。

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好了不多说正式进入题解。

其实楼上的巨佬们都已经讲得很清楚了，就是从1到各个点的最短路和各个点到1的最短路的权值和加起来就行了。

这里给大家安利一下堆优化+\(Dijkstra\)的代码（其实不用加\(int\)其实我是先做了1342再复制粘贴过这题的233）：

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#define ll long long
using namespace std;

ll n, m, u, v, w, h1[2000007], h2[2000007], cnt1, cnt2, dis[2000007], vis[2000007];
struct edge {
	ll v, to, nxt;
}e1[2000007], e2[2000007];
priority_queue<pair<ll, ll> > q;

inline ll read() {
	ll f = 1ll, x = 0ll;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-')	f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		x = x * 10 + c - '0';
		c = getchar();
	}
	return x * f;
}
void a_e(ll u, ll v, ll w) {
	e1[++cnt1] = (edge){w, v, h1[u]}; h1[u] = cnt1;
	e2[++cnt2] = (edge){w, u, h2[v]}; h2[v] = cnt2;
}
ll dj() {
	ll ans = 0;
	memset(dis, 0x3f, sizeof(dis));
	dis[1] = 0;
	q.push(make_pair(0ll, 1ll));
	while(!q.empty()) {
		ll x = q.top().second;
		q.pop();
		if(vis[x])	continue;
		vis[x] = 1;
		for(ll i = h1[x]; i; i = e1[i].nxt) {
			ll y = e1[i].to, z = e1[i].v;
			if(dis[y] > dis[x] + z) {
				dis[y] = dis[x] + z;
				q.push(make_pair(-dis[y], y));
			}
		}
	}
	for(ll i = 2; i <= n; ++i) {
//		printf("1 to %d's distance is %d.\n", i, dis[i]);
		ans += dis[i];
	}
	while(!q.empty())	q.pop();
	memset(dis, 0x3f, sizeof(dis));
	memset(vis, 0, sizeof(vis));
	dis[1] = 0;
	q.push(make_pair(0ll, 1ll));
	while(!q.empty()) {
		ll x = q.top().second;
		q.pop();
		if(vis[x])	continue;
		vis[x] = 1;
		for(ll i = h2[x]; i; i = e2[i].nxt) {
			ll y = e2[i].to, z = e2[i].v;
			if(dis[y] > dis[x] + z) {
				dis[y] = dis[x] + z;
				q.push(make_pair(-dis[y], y));
			}
		}
	}
	for(ll i = 2; i <= n; ++i) {
//		printf("%d to 1's distance is %d.\n", i, dis[i]);
		ans += dis[i];
	}
	return ans;
}

int main() {
	n = read(), m = read();
	for(ll i = 1; i <= m; ++i) {
		u = read(), v = read(), w = read();
		a_e(u, v, w);
	}
	printf("%lld", dj());
	return 0;
}

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