hdu 2448 Mining Station on the Sea(最短路+費用流)

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=2448 

題意：給你一個由N個港口和M個海上油田構成的連通無向圖(給出了圖中所有的邊和權值),現在給你N個船所在的油田編號,問你讓這N條船,每條都回到1個港口去(每個港口最多隻能容納一條船),問你這N條船行走的總距離最短是多少?

解題思路：首先可以用Floyd算法，求出油田到港口的最短路徑。由於一個船對應一個港口，所以這裏容易想到是一個二分圖的模型，把港口放在X集合，船放在Y集合，然後根據船i所在的油田到港口j的最短路徑，連一條容量爲1，費用爲最短路徑的邊。流一遍費用流即可。

PS:注意邊的方向。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;

const int maxn = 1000;
const int inf = 0x3f3f3f3f;
struct Edge
{
	int from,to,flow,cost,next;

	Edge(){}
	Edge(int f,int t,int fl,int co):from(f),to(t),flow(fl),cost(co){} 
};
struct MCMF
{
	int n,s,t;
	vector<Edge> edge;
	vector<int> G[maxn<<1];
	int dis[maxn<<1];
	int pre[maxn<<1];
	bool inq[maxn<<1];

	void init(int n,int s,int t)
	{
		this->n = n, this->s = s, this->t = t;
		edge.clear();
		for(int i = 0; i <= n; i++) G[i].clear();
	}

	void addedge(int u,int v,int flow,int cost)
	{
		edge.push_back(Edge(u,v,flow,cost));
		edge.push_back(Edge(v,u,0,-cost));
		int m = edge.size();
		G[u].push_back(m-2);
		G[v].push_back(m-1);
	}

	int spfa()
	{
		queue<int> q;
		memset(dis,inf,sizeof(dis));
		memset(pre,-1,sizeof(pre));
		memset(inq,false,sizeof(inq));
		dis[s] = 0;
		inq[s] = true;
		q.push(s);
		while(!q.empty())
		{
			int u = q.front();
			q.pop();
			inq[u] = false;
			for(int i = 0; i < G[u].size(); i++)
			{
				int v = edge[G[u][i]].to;
				if(dis[v] > dis[u] + edge[G[u][i]].cost && edge[G[u][i]].flow > 0)
				{
					dis[v] = dis[u] + edge[G[u][i]].cost;
					pre[v] = G[u][i];
					if(inq[v] == false)
					{
						inq[v] = true;
						q.push(v);
					}
				}
			}
		}
		return dis[t] != inf;
	}

	int solve()
	{
		int mincost = 0,minflow;
		while(spfa())
		{
			minflow = inf;
			for(int i = pre[t]; i != -1; i = pre[edge[i].from])
				minflow = min(minflow,edge[i].flow);
			for(int i = pre[t]; i != -1; i = pre[edge[i].from])
			{
				edge[i].flow -= minflow;
				edge[i^1].flow += minflow;
			}
			mincost += dis[t] * minflow;
		}
		return mincost;
	}
}mcmf;
int n,m,k,p;
int d[maxn][maxn],bel[105];
 
void floyd()
{
	for(int k = 1; k <= n + m; k++)
		for(int i = 1; i <= n + m; i++)
			for(int j = 1; j <= n + m; j++)
				d[i][j] = min(d[i][j],d[i][k] + d[k][j]);
}

int main()
{
	int u,v,c;
	while(scanf("%d%d%d%d",&n,&m,&k,&p)!=EOF)
	{
		memset(d,inf,sizeof(d));
		for(int i = 1; i <= n; i++)
		{
			scanf("%d",&bel[i]);
			bel[i] += n;
		}
		for(int i = 1; i <= k; i++)
		{
			scanf("%d%d%d",&u,&v,&c);
			d[n + u][n + v] = d[n + v][n + u] = min(d[n + u][n + v],c);
		}
		for(int i = 1; i <= p; i++)
		{
			scanf("%d%d%d",&u,&v,&c);
			d[n + v][u] = min(d[n + v][u],c);
		}
		floyd();
		int s = 0, t = n + m + 1;
		mcmf.init(n + m + 2,s,t);
		for(int i = 1; i <= n; i++)
		{
			mcmf.addedge(s,i,1,0);
			mcmf.addedge(bel[i],t,1,0);
			for(int j = 1; j <= n; j++)
				mcmf.addedge(j,bel[i],1,d[bel[i]][j]);
		}
		printf("%d\n",mcmf.solve());
	}
	return 0;
}