hdu 1853 费用流

找出若干个环覆盖所有的点，使得总的花费最小

因为每个点只能经过一次，所以很快就可想到拆点求最小费用流

建图：

S->i  费用为0 流量为1

i+n->T同上

若有边u->v

u->v+n 费用为边权，容量为1

最后套套模板求一次最小费用流，如果流量等于n，表示每个点都遍历了一次，输出最小费用即可

View Code

#include <iostream>
#include <algorithm>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <memory.h>
#include <queue>
#include <vector>
#include <cmath>
using namespace std;
int sumFlow;
const int MAXN = 502;
const int MAXM = 10002;
const int INF = 1000000000;
struct Edge
{
    int u, v, cap, cost;
    int next;
}edge[MAXM<<2];
int NE;
int head[MAXN], dist[MAXN], pp[MAXN];
bool vis[MAXN];
void init()
{
    NE = 0;
    memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int cap, int cost)
{
    edge[NE].u = u; edge[NE].v = v; edge[NE].cap = cap; edge[NE].cost = cost;
    edge[NE].next = head[u]; head[u] = NE++;
    edge[NE].u = v; edge[NE].v = u; edge[NE].cap = 0; edge[NE].cost = -cost;
    edge[NE].next = head[v]; head[v] = NE++;
}
bool SPFA(int s, int t, int n)
{
    int i, u, v;
    queue <int> qu;
    memset(vis,false,sizeof(vis));
    memset(pp,-1,sizeof(pp));
    for(i = 0; i <= n; i++) dist[i] = INF;
    vis[s] = true; dist[s] = 0;
    qu.push(s);
    while(!qu.empty())
    {
        u = qu.front(); qu.pop(); vis[u] = false;
        for(i = head[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].v;
            if(edge[i].cap && dist[v] > dist[u] + edge[i].cost)
            {
                dist[v] = dist[u] + edge[i].cost;
                pp[v] = i;
                if(!vis[v])
                {
                    qu.push(v);
                    vis[v] = true;
                }
            }
        }
    }
    if(dist[t] == INF) return false;
    return true;
}
int MCMF(int s, int t, int n) // minCostMaxFlow
{
    int flow = 0; // 总流量
    int i, minflow, mincost;
    mincost = 0;
    while(SPFA(s, t, n))
    {
        minflow = INF + 1;
        for(i = pp[t]; i != -1; i = pp[edge[i].u])
            if(edge[i].cap < minflow)
                minflow = edge[i].cap;
        flow += minflow;
        for(i = pp[t]; i != -1; i = pp[edge[i].u])
        {
            edge[i].cap -= minflow;
            edge[i^1].cap += minflow;
        }
        mincost += dist[t] * minflow;
    }
    sumFlow = flow; // 题目需要流量，用于判断
    return mincost;
}
int main()
{
    int n, m;
    int u, v, c;
    while (scanf("%d%d", &n, &m) != -1)
    {
        init();
        int S = 0;
        int T = n + n + 1;
        for (int i = 1; i <= n; ++i)
        {
            addedge(S, i, 1, 0);
            addedge(i + n, T, 1, 0);
        }
        while (m--)
        {
            scanf("%d%d%d", &u, &v, &c);
            addedge(u, v + n, 1, c);
        }
        int ans = MCMF(S, T, T+1);
        if (sumFlow != n) printf("-1\n");
        else printf("%d\n", ans);
    }
    return 0;
}

另外也可以用二分图匹配中的KM算法来搞

View Code

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MIN(a,b) ((a)<(b)?(a):(b))
#define INF 999999
#define MAX 5110
int n,match[MAX];
bool sx[MAX],sy[MAX];
int lx[MAX],ly[MAX],map[MAX][MAX];
bool path(int u)
{
    sx[u]=true;
    for(int v=0;v<n;v++)
        if(!sy[v]&&lx[u]+ly[v]==map[u][v])
        {
            sy[v]=true;
            if(match[v]==-1||path(match[v]))
            {
                match[v]=u;
                return true;
            }
        }
        return false;
}
int KM(bool truth)//可以不用更改地处理最小或最大权匹配
{
    int i,j;
    if(!truth)
    {
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                map[i][j]=-map[i][j];
    }
    for(i=0;i<n;i++)
    {
        lx[i]=-INF;
        ly[i]=0;
        for(j=0;j<n;j++)
            if(lx[i]<map[i][j])
                lx[i]=map[i][j];
    }
    memset(match,-1,sizeof(match));
    for(int u=0;u<n;u++)
        while(1)
        {
            memset(sx,0,sizeof(sx));
            memset(sy,0,sizeof(sy));
            if(path(u)) break;
            int dmin=INF;
            for(i=0;i<n;i++)
                if(sx[i])
                    for(j=0;j<n;j++)
                        if(!sy[j])
                            dmin=MIN(lx[i]+ly[j]-map[i][j],dmin);
                        for(i=0;i<n;i++)
                        {
                            if(sx[i])
                                lx[i]-=dmin;
                            if(sy[i])
                                ly[i]+=dmin;
                        }
        }
        int sum=0;
        for(j=0;j<n;j++)
            sum+=map[match[j]][j];
        if(!truth)
        {
            sum=-sum;
            for(i=0;i<n;i++)
                for(j=0;j<n;j++)
                    map[i][j]=-map[i][j];
        }
        return sum;
}
void Map_Init(int m,int n)
{
    int x,y,w;
    int i,j;
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            map[i][j]=INF;
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&x,&y,&w);
            if(map[x-1][y-1]>w)
                map[x-1][y-1]=w;
        }
}
int main()
{
    int m,i,j,a,b,c;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        Map_Init(m,n);
        int ans=KM(false);
        if(ans<INF) printf("%d\n",ans);
        else printf("-1\n");
    }
}