POJ 3680 Intervals

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离散化+最大费用最大流...

源点---1 .. 2 ..3 .... n ---汇点 连流量为K,费用为0的边

对于(a , b , w) 连从 a到b容量1费用w的边

4

3 1
1 2 2
2 3 4
3 4 8

3 1
1 3 2
2 3 4
3 4 8

3 1
1 100000 100000
1 2 3
100 200 300

3 2
1 100000 100000
1 150 301
100 200 300

14
12
100000
100301

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

using namespace std;

const int maxn=500;
const int INF=0x3f3f3f3f;

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

int Adj[maxn],Size,N;

void init()
{
    Size=0; memset(Adj,-1,sizeof(Adj));
}

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

void Add_Edge(int u,int v,int cap,int cost)
{
    addedge(u,v,cap,cost);
    addedge(v,u,0,-cost);
}

int dist[maxn],vis[maxn],pre[maxn];

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

int n,m,K;

struct Bian
{
    int u,v,w;
}bian[maxn];

int has[maxn],hn;

int main()
{
    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        init();hn=0;

        scanf("%d%d",&m,&K);
        for(int i=0;i<m;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            bian[i].u=a; bian[i].v=b; bian[i].w=c;
            has[hn++]=a; has[hn++]=b;
        }
        sort(has,has+hn);
        hn=unique(has,has+hn)-has;

        n=hn+2;
        N=n;
        for(int i=1;i<n;i++)
        {
            Add_Edge(i-1,i,K,0);
        }

        for(int i=0;i<m;i++)
        {
            bian[i].u=lower_bound(has,has+hn,bian[i].u)-has+1;
            bian[i].v=lower_bound(has,has+hn,bian[i].v)-has+1;
            Add_Edge(bian[i].u,bian[i].v,1,bian[i].w);
        }

        int flow,cost;
        flow=MinCostMaxFlow(0,n-1,cost);
        printf("%d\n",cost);
    }
	return 0;
}

POJ 3680 Intervals

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原文地址：http://blog.csdn.net/ck_boss/article/details/39586141