C - Heavy Transportation（最短路变形 / 最大生成树）

链接：Heavy Transportation

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

---

题意：

有n个城市（编号1~n），m条道路，每条路有最大承重，求从1号城市到n号城市能运输的最大重量。（任意2个城市间最多只有1条道路直达）

---

最短路解法：

该题可看作最短路的变形，同样要维护一个数组d[]，只是d[]的含义变了。

d[u]：表示 从起点s到u的 可运输重量最大的路径 的 最小承重值。

以dijkstra算法为例：

首先初始化 d[s]=INF;（起点到自身可运输重量无穷大），其余d[]均初始化为0。

每次取未访问结点中d[]最大的结点进行访问（记为u），再以u为踏板对u可达的节点进行优化。

关键是如何优化，这是个难点。

假设有路径 u -> v，承重为w

可以发现如果选择了该路径，那么 d[v] = min(d[u] , w)

因为要d[]尽可能的大，那么优化操作如下：

	if(!vis[v]&&d[v]<min(d[u],w))
	{
		d[v]=min(d[u],w);
	}

AC代码：（堆优化的dijkstra算法）

#include<cstdio>
#include<climits>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstring>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=1010;
struct Node
{
	int num;
	int d;
	Node(int a,int b)
	{
		num=a;
		d=b;	
	}
	friend bool operator < (const Node &a,const Node &b)
	{
		return a.d<b.d;        //这样在优先队列中就是d大的在前
	}
};
struct edge
{
	int to;
	int w;
	edge(int a,int b)
	{
		to=a;
		w=b;
	}
};
vector<edge> G[maxn];
int n,m,d[maxn];
bool vis[maxn];
void djs(int s)
{
	memset(d,0,sizeof(d));
	memset(vis,false,sizeof(vis));
	d[s]=INF;
	priority_queue<Node> q;
	q.push(Node(s,d[s]));
	while(!q.empty())
	{
		int u=q.top().num;
		q.pop();
		if(vis[u])
			continue;
		else
			vis[u]=true;
		for(int i=0;i<G[u].size();i++)
		{
			int v=G[u][i].to,w=G[u][i].w;
			if(!vis[v]&&d[v]<min(d[u],w))
			{
				d[v]=min(d[u],w);
				q.push(Node(v,d[v]));
			}
		}
	}
}
int main()
{	
	int T,kase=0;
	scanf("%d",&T);
	while(T--)
	{
		for(int i=0;i<maxn;i++)
			G[i].clear();
		scanf("%d %d",&n,&m);
		for(int i=1;i<=m;i++)
		{
			int u,v,w;
			scanf("%d %d %d",&u,&v,&w);
			G[u].push_back(edge(v,w));
			G[v].push_back(edge(u,w));
		}
		djs(1);
		printf("Scenario #%d:\n",++kase);
		printf("%d\n\n",d[n]);
	}
	return 0;
}

---

最大生成树解法：

取尽可能大的边将城市连通（构建最大生成树），这样1->n的路径上最小的承重即为答案。

用kruskal算法来求最大生成树（这个解法比dijkstra算法要好理解很多），只是终止条件要改动一下。

kruskal算法求最大生成树，是按边贪心，先将边从大到小排序，再遍历边。
若该边两端点不在同一连通块中，那么就将该边加入最大生成树，并且令两端点的连通块合并；否则跳过该边。

原本的终止条件是当边数 cnt == n-1即终止，而该题是当1号和n号位于同一连通块中即结束，且当前边的边权为答案。

为什么呢？

因为当检测到1号和n号位于同一连通块中，说明当前边令1号和n号所处的2个连通块合并了，那么1->n的路径已经形成，且当前边必是该路径上的一边，且是边权最小的一边。（边是从大到小遍历的）

AC代码：

#include<cstdio>
#include<climits>
#include<algorithm>
#include<vector>
#include<cstring>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=1010;
struct road
{
	int u,v;
	int w;
};
bool cmp(const road &a,const road &b) { return a.w>b.w; }
int n,m;
int father[maxn];
vector<road> a;
void init()
{
	for(int i=0;i<maxn;i++)
		father[i]=i;
}
int findfather(int x)
{
	int root=x;
	while(root!=father[root])
		root=father[root];
		
	//路径压缩
	while(x!=father[x])
	{
		int t=x;
		x=father[x];
		father[t]=root;
	}
	
	return root;
}
int kruskal()
{
	init();
	sort(a.begin(),a.end(),cmp);
	for(int i=0;i<a.size();i++)
	{
		int Fu=findfather(a[i].u);
		int Fv=findfather(a[i].v);
		if(Fu!=Fv)
		{
			father[Fu]=Fv;
			if(findfather(1)==findfather(n))
				return a[i].w;
		}
	}
}
int main()
{	
	int T,kase=0;
	scanf("%d",&T);
	while(T--)
	{
		a.clear();
		scanf("%d %d",&n,&m);
		for(int i=1;i<=m;i++)
		{
			road t;
			scanf("%d %d %d",&t.u,&t.v,&t.w);
			a.push_back(t);
		}
		printf("Scenario #%d:\n",++kase);
		printf("%d\n\n",kruskal());
	}
	return 0;
}