ACM: 最大流 图论题 poj 1797

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本文介绍了一种解决最大流问题的方法，特别关注于寻找从起点到终点的最大运输能力。通过使用SPFA算法进行流量松弛操作，确保每条边的流量不超过其容量限制。

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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

Sample Output

Scenario #1:


4



题意: FJ要从1点运送东西到n点. 要求出最大的运送货物总量. (最大流)

      但是每条路的重量的有限制的.即: 最小中找最大的.



解题思路:

         1. 最大流问题. flow记录源点到当前节点的最大流.

         2. 假设u -> v 的当前流量是cur.

          松弛操作是:  int t = min(flow[u],cur);

                      if(flow[v] < t) flow[v] = t;

代码:
#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#define MAX 200005
using namespace std;
const int INF = (1<<30);

struct node
{
   int u, v;
   int w;
   int next;
}edges[MAX];

int n, m;
int num;
int first[MAX];
int flow[MAX];
bool vis[MAX];

inline int min(int a,int b)
{
   return a > b ? b : a;
}

void read_graph()
{
   num = 0;
   memset(edges,0,sizeof(edges));
   memset(first,-1,sizeof(first));

   int u, v, w;
   for(int i = 0; i < m; ++i)
   {
       scanf("%d %d %d",&u,&v,&w);
       edges[num].u = u;
       edges[num].v = v;
       edges[num].w = w;
       edges[num].next = first[u];
       first[u] = num++;

       edges[num].u = v;
       edges[num].v = u;
       edges[num].w = w;
       edges[num].next = first[v];
       first[v] = num++;
   }
}

int spfa()
{
   memset(vis,false,sizeof(vis));
   memset(flow,0,sizeof(flow));

   queue<int> qu;
   qu.push(1);
   vis[1] = true;
   flow[1] = INF;

   while( !qu.empty() )
   {
       int u = qu.front();
       qu.pop();
       vis[u] = false;

       for(int v = first[u]; v != -1; v = edges[v].next)
       {
           int k = min(flow[u],edges[v].w); //要通过的流量与当前的限制流量比较
           if(flow[edges[v].v] < k) //流量松弛操作
           {
               flow[edges[v].v] = k;
               if( !vis[edges[v].v] )
               {
                   vis[edges[v].v] = true;
                   qu.push(edges[v].v);
               }
           }
       }
   }

   return flow[n];
}

int main()
{
//   freopen("input.txt","r",stdin);
   int t = 1;
   int casesize;
   scanf("%d",&casesize);

   while( casesize-- )
   {
       scanf("%d %d",&n,&m);
       printf("Scenario #%d:\n",t++);
       read_graph();
       printf("%d\n\n",spfa());


   }
   return 0;
}