POJ 1797 Heavy Transportation（最大生成树/最短路变形）

Posted 2020-08-24

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Heavy Transportation

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 31882		Accepted: 8445

Description

Background 
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight. 
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know. 

Problem 
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo‘s place) to crossing n (the customer‘s place). You may assume that there is at least one path. All streets can be travelled in both directions.

Input

The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.

Sample Input

1
3 3
1 2 3
1 3 4
2 3 5

Sample Output

Scenario #1:
4

解题思路

题意：从找一条从 1 到 n 的道路，使得这条道路的每段距离中的最小值最大。

思路：利用Kruskal的思想，把每段道路的权值按照从大到小排序，然后依次加边直至 1 可达 n为止，此时这条路中最小直即为答案。

利用dijkstra或是spfa的思想去解，dis[ x ]表示从 1 到达 x 的道路中，其道路段的最小值的最大值。

 

Kruskal()

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1005;
struct Edge{
	int u,v,w;
}edge[maxn*maxn/2];
int fa[maxn];

bool cmp(Edge a,Edge b)
{
	return a.w > b.w;
}

int find(int x)
{
	return fa[x] == x?fa[x]:fa[x] = find(fa[x]);
}

void Union(int x,int y)
{
	int fx = find(x),fy = find(y);
	if (fx != fy)	fa[fx] = fy;
}

int main()
{
	int tcase;
	scanf("%d",&tcase);
	for (int t = 1;t <= tcase;t++)
	{
		int n,m,res = 0x3f3f3f3f;
		scanf("%d%d",&n,&m);
		for (int i = 0;i <= n;i++)	fa[i] = i;
		for (int i = 0;i < m;i++)
		{
			scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
		}
		sort(edge,edge + m,cmp);
		for (int i = 0; i< m;i++)
		{
			Union(edge[i].u,edge[i].v);
			if (find(1) == find(n))
			{
				res = edge[i].w;
				break;
			}
		}
		printf("Scenario #%d:\n",t);
		printf("%d\n",res);
		if (t != tcase)	printf("\n");
	}
	return 0;
} 

dijkstra（）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 1005;
const int INF = 0x3f3f3f3f;
struct Edge{
	int u,v,w,nxt;
	bool operator < (const Edge &a)const
	{
		return w < a.w;
	}
}edge[maxn*maxn];
int tot = 0,head[maxn],dis[maxn];
bool vis[maxn];

void addedge(int u,int v,int w)
{
	edge[tot] = (Edge){u,v,w,head[u]
	};
	head[u] = tot++;
}

void dijkstra(int st,int n)
{
	Edge p;
	priority_queue<Edge>que;
	memset(dis,0,sizeof(dis));
	memset(vis,false,sizeof(vis));
	p.v = st;
	p.w = INF;
	dis[st] = INF;
	que.push(p);
	while (!que.empty())
	{
		p = que.top();
		que.pop();
		int u = p.v;
		if (vis[u])	continue;
		vis[u] = true;
		for (int i = head[u];~i;i = edge[i].nxt)
		{
			int v = edge[i].v,w = edge[i].w;
			if (dis[v] < min(dis[u],w))
			{
				dis[v] = min(dis[u],w);
				p.v = v,p.w = dis[v];
				que.push(p); 
			}
		}
	}
}


int main()
{
	int tcase;
	scanf("%d",&tcase);
	for (int t = 1;t <= tcase;t++)
	{
		int n,m,u,v,w;
		memset(head,-1,sizeof(head));
		scanf("%d%d",&n,&m);
		for (int i = 0;i < m;i++)
		{
			scanf("%d%d%d",&u,&v,&w);
			addedge(u,v,w);
			addedge(v,u,w);
		}
		dijkstra(1,n);
		printf("Scenario #%d:\n",t);
		printf("%d\n",dis[n]);
		if (t != tcase)	printf("\n");
	}
	return 0;
	
} 

spfa()

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 1005;
const int INF = 0x3f3f3f3f;
struct Edge{
	int u,v,w,nxt;
}edge[maxn*maxn];
int tot = 0,head[maxn],dis[maxn];
bool vis[maxn];

void addedge(int u,int v,int w)
{
	edge[tot] = (Edge){u,v,w,head[u]
	};
	head[u] = tot++;
}


void spfa(int st,int ed)
{
	memset(vis,false,sizeof(vis));
	memset(dis,0,sizeof(dis));
	queue<int>que;
	dis[st] = INF;
	que.push(st);
	vis[st] = true;
	while (!que.empty())
	{
		int u = que.front();
		que.pop();
		vis[u] = false;
		for (int i = head[u];~i;i = edge[i].nxt)
		{
			int v = edge[i].v,w = edge[i].w;
			if (min(dis[u],w) > dis[v])
			{
				dis[v] = min(dis[u],w);
				if (!vis[v])
				{
					que.push(v);
					vis[v] = true;
				}
			}
		}
	}
}

int main()
{
	int tcase;
	scanf("%d",&tcase);
	for (int t = 1;t <= tcase;t++)
	{
		int n,m,u,v,w;
		memset(head,-1,sizeof(head));
		scanf("%d%d",&n,&m);
		for (int i = 0;i < m;i++)
		{
			scanf("%d%d%d",&u,&v,&w);
			addedge(u,v,w);
			addedge(v,u,w);
		} 
		spfa(1,n);
		printf("Scenario #%d:\n",t);
		printf("%d\n",dis[n]);
		if (t != tcase)	printf("\n");
	}
}