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HDU1535/POJ1511 - Invitation Cards - 有向图来回最短路

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1.题目描述：

Invitation Cards

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4061 Accepted Submission(s): 1848

Problem Description

In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.
The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where 'X' denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.

All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.

Input

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case begins with a line containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is the number of stops including CCS and Q the number of bus lines. Then there are Q lines, each describing one bus line. Each of the lines contains exactly three numbers - the originating stop, the destination stop and the price. The CCS is designated by number 1. Prices are positive integers the sum of which is smaller than 1000000000. You can also assume it is always possible to get from any stop to any other stop.

Output

For each case, print one line containing the minimum amount of money to be paid each day by ACM for the travel costs of its volunteers.

Sample Input

Sample Output

Source

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2.题意概述：

给你一个源点，让你从这里派发n个学生去其余的n-1个站点去邀请人们去CSS，然后再返回CSS，使得总的cost最小。

3.解题思路：

这题MLE了好几发，得出一个教训，vector在大数据面前还是没有链式前向星来的好。

回正题，既然是来回，那么就可以考虑正反建图，正图的最短路+返程的最短路就是总的最短路。然后数据还是比较水的，本来觉得会爆long long，结果没有，POJ有题一模一样，但是数据更死，而且会爆ll，但是奇怪的是这个做法在POJ交却过不了。

4.AC代码：

HDU:

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 1000010
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
struct node
{
	int to, val, next;
} edge[maxn];
struct point
{
	int u, v, w;
} edge_[maxn];
int head[maxn], dis[maxn];
int cnt;
void init()
{
	memset(head, -1, sizeof(head));
	cnt = 0;
}
void addedge(int u, int v, int w)
{
	edge[cnt].to = v;
	edge[cnt].val = w;
	edge[cnt].next = head[u];
	head[u] = cnt++;
}
void spfa(int sta, int n)
{
	fill(dis, dis + n + 1, INF);
	queue<int> q;
	dis[sta] = 0;
	q.push(sta);
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].to;
			int w = edge[i].val;
			if (dis[v] > dis[u] + w)
			{
				dis[v] = dis[u] + w;
				q.push(v);
			}
		}
	}
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	long _begin_time = clock();
#endif
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int p, q;
		scanf("%d%d", &p, &q);
		init();
		for (int i = 0; i < q; i++)
		{
			scanf("%d%d%d", &edge_[i].u, &edge_[i].v, &edge_[i].w);
			addedge(edge_[i].u, edge_[i].v, edge_[i].w);
		}
		int ans = 0;
		spfa(1, cnt);
		for (int i = 2; i <= p; i++)
			ans += dis[i];
		init();
		for (int i = 0; i < q; i++)
			addedge(edge_[i].v, edge_[i].u, edge_[i].w);
		spfa(1, cnt);
		for (int i = 2; i <= p; i++)
			ans += dis[i];
		printf("%d\n", ans);
	}
#ifndef ONLINE_JUDGE
	long _end_time = clock();
	printf("time = %ld ms.", _end_time - _begin_time);
#endif
	return 0;
}

POJ:

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <ctime>
#define INF 0x7fffffff
#define maxn 1000010
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
struct node
{
	int to, next;
	int val;
} edge[maxn];
struct point
{
	int u, v;
	int w;
} edge_[maxn];
int head[maxn];
int dis[maxn];
bool vis[maxn];
int cnt;
void init()
{
	memset(head, -1, sizeof(head));
	cnt = 0;
}
void addedge(int u, int v, int w)
{
	edge[cnt].to = v;
	edge[cnt].val = w;
	edge[cnt].next = head[u];
	head[u] = cnt++;
}
void spfa(int sta, int n)
{
	memset(vis, 0, sizeof(vis));
	fill(dis, dis + n + 1, INF);
	deque<int> q;
	vis[sta] = 1;
	dis[sta] = 0;
	q.push_back(sta);
	while (!q.empty())
	{
		int u = q.front();
		q.pop_front();
		vis[u] = 0;
		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].to;
			int w = edge[i].val;
			if (dis[v] > dis[u] + w)
			{
				dis[v] = dis[u] + w;
				if (!vis[v])
				{
					vis[v] = 1;
					if (!q.empty() && w <= dis[q.front()])
						q.push_front(v);
					else
						q.push_back(v);
				}
			}
		}
	}
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	long _begin_time = clock();
#endif
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int p, q;
		scanf("%d%d", &p, &q);
		init();
		for (int i = 0; i < q; i++)
		{
			scanf("%d%d%d", &edge_[i].u, &edge_[i].v, &edge_[i].w);
			addedge(edge_[i].u, edge_[i].v, edge_[i].w);
		}
		ll ans = 0;
		spfa(1, cnt);
		for (int i = 2; i <= p; i++)
			ans += dis[i];
		init();
		for (int i = 0; i < q; i++)
			addedge(edge_[i].v, edge_[i].u, edge_[i].w);
		spfa(1, cnt);
		for (int i = 2; i <= p; i++)
			ans += dis[i];
		printf("%lld\n", ans);
	}
#ifndef ONLINE_JUDGE
	long _end_time = clock();
	printf("time = %ld ms.", _end_time - _begin_time);
#endif
	return 0;
}

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