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标签：最小费用最大流

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题意：在一个王国有N个城市，M条路。选择N条路，构成一个环路。求出最小距离。

解析：构图，最小费用最大流。将源点和终点至个点花费记作0，然后将所有路径流量记作1或同一值。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;


const int maxn = 10000;
const int maxm = 100000;
const int INF = 0xfffffff;

struct Edge{
	int to, next, cap, flow, cost;
}edge[ maxm ];

int head[ maxn ], tol;
int pre[ maxn ], dis[ maxn ];
bool vis[ maxn ];

int N;

void init( int n ){
	N = n;
	tol = 0;
	memset( head, -1, sizeof( head ) );
}

void addedge( int u, int v, int cap, int cost ){
	edge[ tol ].to = v;
	edge[ tol ].cap = cap;
	edge[ tol ].cost = cost;
	edge[ tol ].flow = 0;
	edge[ tol ].next = head[ u ];
	head[ u ] = tol++;
	edge[ tol ].to = u;
	edge[ tol ].cap = 0;
	edge[ tol ].cost = -cost;
	edge[ tol ].flow = 0;
	edge[ tol ].next = head[ v ];
	head[ v ] = tol++;
}

bool spfa( int s, int t ){
	queue< int > q;
	for( int i = 0; i < N; ++i ){
		dis[ i ] = INF;
		vis[ i ] = false;
		pre[ i ] = -1;
	}
	dis[ s ] = 0;
	vis[ s ] = true;
	q.push( s );
	while( !q.empty( ) ){
		int u = q.front();
		q.pop();
		vis[ u ] = false;
		for( int i = head[ u ]; i != - 1; i = edge[ i ].next ){
			int v = edge[ i ].to;
			if( edge[ i ].cap > edge[ i ].flow && dis[ v ] > dis[ u ] + edge[ i ].cost ){
				dis[ v ] = dis[ u ] + edge[ i ].cost;
				pre[ v ] = i;
				if( !vis[ v ] ){
					vis[ v ] = true;
					q.push( v );
				}
			}
		}
	}
	if( pre[ t ] == -1 ) 
		return false;
	else
		return true;
}

int minCostMaxflow( int s, int t, int &cost ){
	int flow = 0;
	cost = 0;
	while( spfa( s, t ) ){
		int Min = INF;
		for( int i = pre[ t ]; i != - 1; i = pre[ edge[ i ^ 1 ].to ] ){
			if( Min > edge[ i ].cap - edge[ i ].flow )
				Min = edge[ i ].cap - edge[ i ].flow;
		}
		for( int i = pre[ t ]; i  != -1; i = pre[ edge[ i ^ 1 ].to ] ){
			edge[ i ].flow += Min;
			edge[ i ^ 1 ].flow -= Min;
			cost += edge[ i ].cost * Min;
		}
		flow += Min;
	}
	return cost;
}

int main(){
	int Case;
	int n, m;
	scanf( "%d", &Case );
	while( Case-- ){
		scanf( "%d%d", &n, &m );
		int N = 2 * n + 2;
		int start = 0, end = 2 * n + 1;
		init( N );
		for( int i = 1; i <= n; ++ i ){
			addedge( 0, i, 1, 0 );
		}
		for( int i = 1; i <= n; ++i ){
			addedge( n + i, end, 1, 0 );
		}
		int x, y, value;
		for( int i = 0; i < m; ++i ){
			scanf( "%d%d%d", &x, &y, &value );
			addedge( x, n + y, 1, value );
		}
		int cost;
		int ans = minCostMaxflow( start, end, cost );
		cout << ans << endl;
		
	}
	return 0;
}

Tour

标签：最小费用最大流

原文地址：http://blog.csdn.net/bo_jwolf/article/details/38022447