poj 3680 Intervals(费用流)

http://poj.org/problem?id=3680

巧妙的构图。

题目：给定N个区间（ai，bi）权值wi，求最大权和且每个点最多覆盖K次。

构图：将区间端点离散化，将第i个点连第i+1个点花费为0，容量为INF，即addedge(i,i+1,0,INF)（可用来跳过一些区间）;

　　　再处理N个区间（ai，bi），addedge(ai,bi,-wi,1);

　　　最后源点连第一个点，addedge(src,1,0,k);最后一个点连汇点,addedge(n,sink,0,k)。

原理：构完图之后做最小费用最大流，其实就是找不大于K条增广路。如果两个区间有交集，则不能一次流过，要分成两条增广路流过；反之，则可一次流过。

/*
 *Author:       Zhaofa Fang
 *Created time: 2013-07-19-10.41
 *Language:     C++
 */
#include <cstdio>
#include <cstdlib>
#include <sstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;

typedef long long ll;
typedef pair<int,int> PII;
#define DEBUG(x) cout<< #x << ':' << x << endl
#define FOR(i,s,t) for(int i = (s);i <= (t);i++)
#define FORD(i,s,t) for(int i = (s);i >= (t);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define REPD(i,n) for(int i=(n-1);i>=0;i--)
#define PII pair<int,int>
#define PB push_back
#define MP make_pair
#define ft first
#define sd second
#define lowbit(x) (x&(-x))
#define INF (1<<30)
#define eps (1e-8)

const int maxm = 5000;
const int maxn = 405;
int ans,anscost;
struct Edge {
    int u,v,cost,cap,flow,next;
}et[maxm];
int S,T;
int eh[maxn],tot,low[maxn],p[maxn],dist[maxn];
bool vist[maxn];
bool spfa(){
    queue<int>que;
    memset(vist,0,sizeof(vist));
    memset(p,-1,sizeof(p));
    fill(dist,dist+maxn,INF);
    vist[S] = 1,low[S] = INF,dist[S] = 0;
    que.push(S);
    while(!que.empty()){
        int u = que.front();
        que.pop();
        vist[u] = false;
        for(int i=eh[u];i!=-1;i=et[i].next){
            int v = et[i].v,cost = et[i].cost,cap=et[i].cap,flow=et[i].flow;
            if(flow < cap && dist[u] + cost < dist[v]){
                dist[v] = dist[u] + cost;
                p[v] = i;
                low[v] = min(low[u],cap-flow);
                if(!vist[v]){
                    vist[v] = 1;
                    que.push(v);
                }
            }
        }
    }
    return dist[T]!=INF;
}
void costflow(){
    ans = 0,anscost = 0;
    int num=0;
    while(spfa()){
        int x = p[T];
        ans += low[T];
        anscost += low[T]*dist[T];
        while(x!=-1){
            et[x].flow += low[T];
            et[x^1].flow -= low[T];
            et[x^1].cost = -et[x].cost;
            x = p[et[x].u];
        }
    }
}
void add(int u,int v,int cost,int cap,int flow){
    Edge e = {u,v,cost,cap,flow,eh[u]};
    et[tot] = e;
    eh[u] = tot ++;
}
void addedge(int u,int v,int cost,int cap){
    add(u,v,cost,cap,0);add(v,u,-cost,0,0);
}
void init(){
    tot = 0;
    memset(eh,-1,sizeof(eh));
}
int X[maxn];
int a[maxn],b[maxn],w[maxn];
int find(int key,int n){
    int r = lower_bound(X,X+n,key)-X;
    return r+1;
}
int main(){
    //freopen("in","r",stdin);
    //freopen("out","w",stdout);
    int Ca;
    scanf("%d",&Ca);
    while(Ca--){
        int N,K;
        scanf("%d%d",&N,&K);
        int cnt = 0;
        init();
        REP(i,N){
            scanf("%d%d%d",&a[i],&b[i],&w[i]);
            X[cnt++]=a[i];X[cnt++]=b[i];
            w[i] = -w[i];
        }
        sort(X,X+cnt);
        int n = 1;
        FOR(i,1,cnt-1){
            if(X[i]!=X[i-1])X[n++]=X[i];
        }
        REP(i,n-1){
            addedge(i+1,i+2,0,INF);
        }
        REP(i,N){
            int l = find(a[i],n);
            int r = find(b[i],n);
            addedge(l,r,w[i],1);
        }
        addedge(0,1,0,K);
        addedge(n,n+1,0,K);
        S = 0;T = n+1;costflow();
        printf("%d
",-anscost);

    }
    return 0;
}