POJ 1797 Heavy Transportation（二分+宽搜）

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

题意：你有一辆卡车可以在城市之间运输，城市与城市之间有一个最大载重量。现在要你计算从城市1到城市n卡车最大载重量是多少。

题解：找一条路使得最小的道路载重量尽可能大。
二分+bfs
注意：每次输出答案后要输出一行空行。

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int maxn=1e6+10;
struct cc{
    int from,to,cost;
}es[maxn];
int first[maxn],nxt[maxn];
bool vis[1000+10];
int tot=0;
void build(int ff,int tt,int pp)
{
    es[++tot]=(cc){ff,tt,pp};
    nxt[tot]=first[ff];
    first[ff]=tot;
}
int n,m;
queue<int>q;
bool bfs(int mid)
{
    q.push(1);
    vis[1]=1;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        if(u==n)
        {
            return 1;
        }
        for(int i=first[u];i;i=nxt[i])
        {
            int v=es[i].to;
            if(es[i].cost>=mid)
            {
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
    return 0;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int k=1;k<=T;k++)
    {
        tot=0;
        memset(first,0,sizeof(first));
        memset(nxt,0,sizeof(nxt));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            build(x,y,z);
            build(y,x,z);
        }
        int l=0,r=1e9;
        int ans=0;
        while(l<=r)
        {
            int mid=(l+r)/2;
            memset(vis,0,sizeof(vis));
            while(!q.empty())
            {
                q.pop();
            }
            if(bfs(mid))
            {
                l=mid+1;
                ans=max(ans,mid);
            }
            else
            {
                r=mid-1;
            }
        }
        printf("Scenario #%d:\n",k);
        printf("%d\n",ans);
        printf("\n");
    }
    return 0;
}