POJ3680-Intervals

Intervals

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 8363 Accepted: 3569

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 

题意：给出n个区间及其权值，从中选出一些区间使它们的权值合最大，任意一个点最多可以被覆盖k次
解题思路：先将所有点离散化一下，把区间所有的端点从小到大编号构图：
1.连接第i和i + 1个端点，容量设置为k，权值为0
2.连接第一个端点和源点，连接最后的端点和汇点，容量为k，权值为0（这样可以使每个点覆盖不超过k次）
3.连接所有区间端点（左边的连向右边的），容量为1，权值为该区间的权值*（ - 1）（一个区间只能使用一次）
跑一遍最小费用流得出答案

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int vis[405],d[405],pre[405],a[405];struct Edge{    int u, v, c, cost, next;} edge[200009];int s[405], cnt;void init(){    cnt = 0;    memset(s, -1, sizeof(s));}void add(int u, int v, int c, int cost){    edge[cnt].u = u;    edge[cnt].v = v;    edge[cnt].cost = cost;    edge[cnt].c = c;    edge[cnt].next = s[u];    s[u] = cnt++;    edge[cnt].u = v;    edge[cnt].v = u;    edge[cnt].cost = -cost;    edge[cnt].c = 0;    edge[cnt].next = s[v];    s[v] = cnt++;}bool spfa(int ss, int ee,int &flow,int &cost){    queue<int> q;    memset(d, INF, sizeof d);    memset(vis, 0, sizeof vis);    d[ss] = 0, vis[ss] = 1, pre[ss] = 0, a[ss] = INF;    q.push(ss);    while (!q.empty())    {        int u = q.front();q.pop();        vis[u] = 0;        for (int i = s[u]; ~i; i = edge[i].next)        {            int v = edge[i].v;            if (edge[i].c>0&& d[v]>d[u] + edge[i].cost)            {                d[v] = d[u] + edge[i].cost;                pre[v] = i;                a[v] = min(a[u], edge[i].c);                if (!vis[v])                {                    vis[v] = 1;                    q.push(v);                }            }        }    }    if (d[ee] == INF) return 0;    flow += a[ee];    cost += d[ee]*a[ee];    int u = ee;    while (u != ss)    {        edge[pre[u]].c -= a[ee];        edge[pre[u] ^ 1].c += a[ee];        u = edge[pre[u]].u;    }    return 1;}int MCMF(int ss, int ee){    int cost = 0, flow=0;    while (spfa(ss, ee, flow, cost));    return cost;}int l[205],r[205],val[205],x[405];int main(){    int n,k,t;    scanf("%d",&t);    while (t--)    {        scanf("%d%d", &n, &k);        init();        for(int i=1;i<=n;i++)        {            scanf("%d%d%d",&l[i],&r[i],&val[i]);            x[i*2-1]=l[i],x[2*i]=r[i];        }        sort(x+1,x+2*n+1);        int m=unique(x+1,x+1+2*n)-x;        int s = 0, e = m ;        for (int i = 1; i < m-1; i++) add(i, i+1, k, 0);        add(s,1,k,0);add(m-1,e,k,0);        for(int i=1;i<=n;i++)        {            int ll=lower_bound(x+1,x+m,l[i])-x;            int rr=lower_bound(x+1,x+m,r[i])-x;            add(ll,rr,1,-val[i]);        }        printf("%d\n", -MCMF(s, e));    }    return 0;}