BZOJ 1738: [Usaco2005 mar]Ombrophobic Bovines 发抖的牛网络流_二分_Floyd

最新推荐文章于 2019-06-04 11:04:00 发布

原创最新推荐文章于 2019-06-04 11:04:00 发布 · 174 阅读

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本文详细解析了BZOJ1738题目——发抖的牛(Ombrophobic Bovines)，通过网络流、二分搜索和Floyd算法解决雨天牛群避雨的复杂问题。代码示例展示了Dinic算法实现最大流，最终求得最小成本。

BZOJ 1738: [Usaco2005 mar]Ombrophobic Bovines 发抖的牛网络流_二分_Floyd

Code:

#include<bits/stdc++.h>
#define setIO(s) freopen(s".in","r",stdin) 
#define maxn 1000000 
#define inf 10000000000000 
#define ll long long 
using namespace std;
namespace Dinic{
    struct Edge{
        int from,to,cap; 
        Edge(int u=0,int v=0,int c=0):from(u),to(v),cap(c){} 
    }; 
    vector<int>G[500]; 
    vector<Edge>edges; 
    queue<int>Q; 
    int vis[500],d[500],curr[500]; 
    int s,t; 
    void addedge(int u,int v,int c){
        edges.push_back(Edge(u,v,c)),edges.push_back(Edge(v,u,0)); 
        int m=edges.size();
        G[u].push_back(m-2),G[v].push_back(m-1); 
    }
    int BFS(){
        memset(vis,0,sizeof(vis)); 
        d[s]=0,vis[s]=1, Q.push(s); 
        while(!Q.empty()){
            int u=Q.front();Q.pop(); 
            for(int sz=G[u].size(),i=0;i<sz;++i){
                Edge r=edges[G[u][i]]; 
                if(!vis[r.to]&&r.cap>0) {
                    vis[r.to]=1,d[r.to]=d[u]+1; 
                    Q.push(r.to); 
                }
            }
        }
        return vis[t]; 
    }
    int dfs(int x,int cur){
        if(x==t) return cur;
        int f,flow=0;
        for(int sz=G[x].size(),i=curr[x];i<sz;++i){
            curr[x]=i; 
            Edge r=edges[G[x][i]]; 
            if(d[r.to]==d[x]+1&&r.cap>0){
                f=dfs(r.to,min(cur,r.cap)); 
                cur-=f,flow+=f,edges[G[x][i]].cap-=f,edges[G[x][i]^1].cap+=f; 
            }
            if(cur<=0) break; 
        }
        return flow; 
    }
    int maxflow(){
        int flow=0;
        while(BFS()) memset(curr,0,sizeof(curr)),flow+=dfs(s,10000000); 
        return flow; 
    }
    void re(){
        for(int i=0;i<500;++i) G[i].clear(); 
        edges.clear(); 
    }
};   
#define row1(i) (i) 
#define row2(i) (i+n) 
int C[maxn],num[maxn],sums=0,n; 
long long d[500][500]; 
bool check(ll tmp)
{
	Dinic::re(); 
	int s=0,t=row2(n+1); 
	Dinic::s=s,Dinic::t=t; 
	for(int i=1;i<=n;++i) 
	{
		if(num[i]) Dinic::addedge(s,row1(i),num[i]);
		if(C[i]) Dinic::addedge(row2(i),t,C[i]);   
	} 
	for(int i=1;i<=n;++i) 
		for(int j=1;j<=n;++j)
		{
			if(i!=j && d[i][j]<=tmp) Dinic::addedge(row1(i), row2(j), 10000000); 
		}
	return Dinic::maxflow() >= sums; 
}
int main()
{
	// setIO("input"); 
	int m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i) scanf("%d%d",&num[i],&C[i]),sums+=num[i];  
	for(int i=0;i<=230;++i) 
	    for(int j=0;j<=230;++j) d[i][j]=inf; 
	for(int i=0;i<=230;++i) d[i][i]=0; 
	for(int i=1;i<=m;++i)
	{
		int u,v;
		ll c;  
		scanf("%d%d%lld",&u,&v,&c); 
		if(u!=v) d[u][v]=d[v][u]=min(d[u][v],c);        
	}
	for(int k=1;k<=n;++k) 
		for(int i=1;i<=n;++i) 
			for(int j=1;j<=n;++j) d[i][j]=min(d[i][j],d[i][k]+d[k][j]); 
	ll l=0,r=100000000000000,ans=-1;  
    while(l<=r)
    {
    	ll mid=(l+r)>>1;
    	if(check(mid)) ans=mid,r=mid-1;
    	else l=mid+1; 
    }
    printf("%lld\n",ans); 
	return 0; 
}