hdu 1532 Drainage Ditches
2015-07-24 23:46
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Drainage Ditches
题意:给出n条水管,m个节点(0~n-1),以及每天水管的最大流量,问从节点0到节点n-1某时刻的最大流量思路:最大流问题
解题方法:初学,参考大神的博客题解,详细点这里
我的理解就是从开始节点开始通过dfs找到去终点的路,然后会发现中途会有交叉口,这时你先选一条继续走,直到找到终点。 注意:当找到终点你会发现之前的流量可能有些是没有用完的,所以你这时候就需要有个反相弧记录多余的流量,然后多余的流量就可以在交叉口继续用了....,每次把到终点的流量值加起来,当最终返回的流量值为0就结束
参考代码:
#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #include<vector> using namespace std; #define N 210 #define INF 0x3f3f3f3f struct Node{ int to,cap,rev; Node(int t,int c,int r){ to=t; cap=c; rev=r; } }; vector<Node>edges ; bool vis ; void init(int n){ for(int i=0;i<=n;i++){ edges[i].clear(); vis[i]=false; } } void add_edge(int from,int to,int cap){ edges[from].push_back(Node(to,cap,edges[to].size())); edges[to].push_back(Node(from,0,edges[from].size()-1)); } int dfs(int s,int t,int f){ if(s==t) return f; vis[s]=true; for(int i=0;i<edges[s].size();i++){ Node &e = edges[s][i]; if(!vis[e.to]&&e.cap>0){ int tmp = dfs(e.to,t,min(f,e.cap)); if(tmp>0){ e.cap-=tmp; edges[e.to][e.rev].cap+=tmp; return tmp; } } } return 0; } int max_flow(int s,int t){ int flow = 0,f=dfs(s,t,INF); while(f){ flow+=f; memset(vis,false,sizeof(vis)); f=dfs(s,t,INF); } return flow; } int main(){ int n,m,a,b,c; while(scanf("%d%d",&n,&m)!=EOF){ init(m); while(n--){ scanf("%d%d%d",&a,&b,&c); add_edge(a,b,c); } printf("%d\n",max_flow(1,m)); } return 0; }
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