poj Heavy Transportation（dijkstra，spfa）

 Heavy Transportation

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

13 31 2 31 3 42 3 5

Sample Output

Scenario #1:4

题意：有n个城市，m个街道，要把货物从1送到n，求1->n的所有路径中能运送货物的最大重量，也就是城市1到城市n每一条通路的最大承载量中的最大值

比如说样例从1到3有两条路，

第一条路1~2~3，1~2的最大承载量为3，2~3的最大承载量为5，那么这条路的最大承载量则为3.

第二条路1~3，1~3的最大承载量为4，那么这条路的最大承载量则为4。

则最大值为4

dijkstra代码：

#include<stdio.h>#include<string.h>#define min(a,b) (a<b?a:b)#define maxn 2000+10int d[maxn],map[maxn][maxn];bool vis[maxn];int n,m;void dijkstra(int st,int en){    int i,j,v;    for(i=1; i<=n; i++)    {        vis[i]=0;        d[i]=map[st][i];//d表示的是i点到起点的最大承载量    }    vis[1]=1;    for(i=1; i<=n-1; i++)    {        int maxx=0;        for(j=1; j<=n; j++)            if(!vis[j]&&d[j]>maxx)                maxx=d[j],v=j;        if(!maxx)            break;        vis[v]=1;        for(j=1; j<=n; j++)            if(!vis[j]&&d[j]<min(d[v],map[v][j]))                d[j]=min(d[v],map[v][j]);    }    printf("%d\n",d[en]);}int main(){    int t,casr;    scanf("%d",&t);    for(casr=1; casr<=t; casr++)    {        scanf("%d%d",&n,&m);        memset(map,0,sizeof(map));        int x,y,z;        while(m--)        {            scanf("%d%d%d",&x,&y,&z);            map[x][y]=map[y][x]=z;        }        printf("Scenario #%d:\n",casr);        dijkstra(1,n);        puts("");    }    return 0;}

spfa代码：

#include<stdio.h>#include<string.h>#include<queue>using namespace std;#define min(a,b) (a<b?a:b)#define N 1000+10#define M 100000+10const int inf=0x3f3f3f3f;int d[N],first[N],next[M];//注意next的大小，next的大小和len（也就是和m）有关bool vis[N];int n,m,len;struct node{    int u,v,w;}G[M];void spfa(int st,int en){    int i;    memset(d,0,sizeof(d));//d表示的是最大承载量    memset(vis,0,sizeof(vis));    vis[st]=1;    d[st]=inf;//起点到起点要置为较大值，使得st为1时，min(d[st],G[i].w)得到正确结果    queue<int>q;    q.push(st);    while(!q.empty())    {        st=q.front();        q.pop();        vis[st]=0;        for(i=first[st];i!=-1;i=next[i])        {            if(d[G[i].v]<min(d[st],G[i].w))            {                d[G[i].v]=min(d[st],G[i].w);                if(!vis[G[i].v])                {                    vis[G[i].v]=1;                    q.push(G[i].v);                }            }        }    }    printf("%d\n",d[en]);}void add_egde(int u,int v,int w)//邻接表存图{    G[len].u=u,G[len].v=v,G[len].w=w;    next[len]=first[u];    first[u]=len++;}int main(){    int t,star;    scanf("%d",&t);    for(star=1;star<=t;star++)    {        scanf("%d%d",&n,&m);        memset(first,-1,sizeof(first));        int u,v,w,i;        len=1;        for(i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&w);            add_egde(u,v,w);            add_egde(v,u,w);        }        printf("Scenario #%d:\n",star);        spfa(1,n);        puts("");    }    return 0;}