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 = ;
 const int maxn = ;
 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,,sizeof(vist));
     memset(p,-,sizeof(p));
     fill(dist,dist+maxn,INF);
     vist[S] = ,low[S] = INF,dist[S] = ;
     que.push(S);
     while(!que.empty()){
         int u = que.front();
         que.pop();
         vist[u] = false;
         for(int i=eh[u];i!=-;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] = ;
                     que.push(v);
                 }
             }
         }
     }
     return dist[T]!=INF;
 }
 void costflow(){
     ans = ,anscost = ;
     int num=;
     while(spfa()){
         int x = p[T];
         ans += low[T];
         anscost += low[T]*dist[T];
         while(x!=-){
             et[x].flow += low[T];
             et[x^].flow -= low[T];
             et[x^].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,);add(v,u,-cost,,);
 }
 void init(){
     tot = ;
     memset(eh,-,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+;
 }
 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 = ;
         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 = ;
         FOR(i,,cnt-){
             if(X[i]!=X[i-])X[n++]=X[i];
         }
         REP(i,n-){
             addedge(i+,i+,,INF);
         }
         REP(i,N){
             int l = find(a[i],n);
             int r = find(b[i],n);
             addedge(l,r,w[i],);
         }
         addedge(,,,K);
         addedge(n,n+,,K);
         S = ;T = n+;costflow();
         printf("%d\n",-anscost);

     }
     return ;
 }

1300+MS