Poj 3680 intervals

Discretization + maximum cost flow...

Source point --- 1... 2... 3... N --- edge with a connection traffic of K and a fee of 0

For edges (a, B, W) that are connected from A to B with a capacity of 1 W

Intervals Time limit:5000 Ms   Memory limit:65536 K Total submissions:6503   Accepted:2691 Description You are givenNWeighted open intervals.ITh interval covers (AI,Bi) And weighsWi. Your task is to pick some of the intervals to maximize the total weights under the limit that no point in the real axis is covered moreKTimes. Input The first line of input is the number of test case. The first line of each test case contains two integers,NAndK(1 ≤K≤N≤ 200 ). The nextNLine each contain three IntegersAI,Bi,Wi(1 ≤AI<Bi≤ 100,000, 1 ≤Wi≤ 100,000) describing the intervals. There is a blank line before each test case. Output For each test case output the maximum total weights in a separate line. Sample Input 43 11 2 22 3 43 4 83 11 3 22 3 43 4 83 11 100000 1000001 2 3100 200 3003 21 100000 1000001 150 301100 200 300 Sample output 1412100000100301 Source Poj founder monthly contest-2008.07.27, windy7926778

#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