POJ3680 Intervals（费用流）

Intervals

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 7564 Accepted: 3172

Description

You are given N weighted open intervals. The ith interval covers (a_i, b_i) and weighs w_i. 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 more than k times.

Input

The first line of input is the number of test case.
The first line of each test case contains two integers, N and K (1 ≤ K ≤ N ≤ 200).
The next N line each contain three integers a_i, b_i, w_i(1 ≤ a_i < b_i ≤ 100,000, 1 ≤ w_i ≤ 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 <stdio.h>#include <algorithm>#include <iostream>using namespace std;const int MAXN = 10000;const int MAXM = 100000;const int INF = 0x3f3f3f3f;struct Edge{    int to,next,cap,flow,cost;}edge[MAXM];int head[MAXN],tol;int pre[MAXN],dis[MAXN];bool vis[MAXN];int N;//节点总个数，节点编号从0~N-1void init(int n){    N = n;    tol = 0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int cap,int cost){    edge[tol].to = v;    edge[tol].cap = cap;    edge[tol].cost = cost;    edge[tol].flow = 0;    edge[tol].next = head[u];    head[u] = tol++;    edge[tol].to = u;    edge[tol].cap = 0;    edge[tol].cost = -cost;    edge[tol].flow = 0;    edge[tol].next = head[v];    head[v] = tol++;}bool spfa(int s,int t){    queue<int>q;    for(int i = 0;i < N;i++)    {        dis[i] = INF;        vis[i] = false;        pre[i] = -1;    }    dis[s] = 0;    vis[s] = true;    q.push(s);    while(!q.empty())    {        int u = q.front();        q.pop();        vis[u] = false;        for(int i = head[u]; i != -1;i = edge[i].next)        {            int v = edge[i].to;            if(edge[i].cap > edge[i].flow &&               dis[v] > dis[u] + edge[i].cost )            {                dis[v] = dis[u] + edge[i].cost;                pre[v] = i;                if(!vis[v])                {                    vis[v] = true;                    q.push(v);                }            }        }    }    if(pre[t] == -1)return false;    else return true;}//返回的是最大流，cost存的是最小费用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 != -1;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 != -1;i = pre[edge[i^1].to])        {            edge[i].flow += Min;            edge[i^1].flow -= Min;            cost += edge[i].cost * Min;        }        flow += Min;    }    return flow;}pair<int,int>p[220];int main(){    int T;    int n,k;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&k);        init(2*n+3);        int a,b,w;        for(int i = 1;i <= n;i++)        {            scanf("%d%d%d",&a,&b,&w);            p[i] = make_pair(a,b);            addedge(2*i-1,2*i,1,-w);            addedge(2*i-1,2*i,INF,0);            addedge(0,2*i-1,1,0);            addedge(2*i,2*n+2,1,0);        }        addedge(2*n+1,0,k,0);        for(int i = 1;i <= n;i++)            for(int j = 1;j <= n;j++)                if(p[j].first >= p[i].second )                    addedge(2*i,2*j-1,INF,0);        int cost = 0;        minCostMaxflow(2*n+1,2*n+2,cost);        printf("%d\n",-cost);    }    return 0;}