最短路径|POJ 1797 Heavy Transportation 最短路径|图

Heavy Transportation

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 38335		Accepted: 10090

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：
无向图，权值代表两个城市的道路所能承受的最大载重量，询问城市1到城市N可以通过的最大载重量是多少？
题目解析：
最短路的变形，Floyd算法指定超时。可以采用Dijkstra算法或者SPFA算法，只是在数组的初始化和判断条件上有变化。首先用一个data数组存两两道路的载重量，不存在的初始化为0，用Distance数组表示城市1到各个城市的最大载重，在跑SPFA算法时初始的Distance[1]设为无穷大，用pin数组作为标记判断是否在队列中。Distance[m]表示城市1到m的最大载重量，data[m][i]代表城市m到i的最大载重量，Distance[i]表示城市1到i的最大载重量，如果Distance[m]与data[m][i]均大于Distance[i],则更新Distacne[i]为两者中的较小者。
【最短路径|POJ 1797 Heavy Transportation】AC代码：
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int MAXN=1e3+10;
const int maxx=INT_MAX;
bool vis[MAXN];
int Distance[MAXN];
int N, M;
int data[MAXN][MAXN];
void SPFA()
{
queueQ;
Distance[1]=maxx;
vis[1]=true;
Q.push(1);
while(!Q.empty())
{
int m=Q.front();
Q.pop();
vis[m]=false;
for(int i=1; i<=N; i++)
{
if(min(Distance[m],data[m][i])>Distance[i])
{
Distance[i]=min(Distance[m],data[m][i]);
if(!vis[i])
{
Q.push(i);
vis[i]=true;
}
}
}
}
}
int main()
{
int T;
while(cin>>T)
{
for(int t=1; t<=T; t++)
{
cin>>N>>M;
memset(data,0,sizeof(data));
memset(vis,false,sizeof(vis));
memset(Distance,0,sizeof(Distance));
int a, b, c;
while(M--)
{
scanf("%d%d%d",&a,&b,&c); ///一定要用scanf读，cin极易超时。
data[a][b]=data[b][a]=c;
}
SPFA();
cout<<"Scenario #"<cout<}
}
return 0;
}