POJ3680 Intervals

Intervals

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 8343 Accepted: 3562

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 

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题目大意：给你n个区间（ai，bi），对应区间的权重为ci。现在让你挑选一些区间，使得任一点出现的次数不超过k，问你满足条件的方案的最大权重和为多少？

思路：最小费用最大流，先拿源点和每个左端点连边，流量为1.费用为0，汇点和右端点连边，流量为1，费用为0.区间左端点和右端点连边费用为-价值，流量为1，然后n2判区间是否重叠，不重叠就连一条费用0，流量无穷的边

#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;#define MAXN 100100#define MAXM 1000100int vis[MAXN],d[MAXN],pre[MAXN],aa[MAXN];struct Edge{    int u, v, c, cost, next;} edge[MAXM];int s[MAXN], 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, aa[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;               aa[v] = min(aa[u], edge[i].c);                if (!vis[v])                {                    vis[v] = 1;                    q.push(v);                }            }        }    }    if (d[ee] == INF) return 0;    flow += aa[ee];    cost += d[ee]*aa[ee];    int u = ee;    while (u != ss)    {        edge[pre[u]].c -= aa[ee];        edge[pre[u] ^ 1].c += aa[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 a[300],b[300];int main(){    int T,m,n,k,u,v,c;    for(scanf("%d",&T);T--;)    {        init();        scanf("%d%d",&n,&k);        add(100001,100002,k,0);        add(100003,100004,k,0);        for(int i=0;i<n;i++)        {            scanf("%d%d%d",&a[i],&b[i],&c);            add(a[i],b[i],1,-c);            add(100002,a[i],1,0);            add(b[i],100003,1,0);        }        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)        {            if(b[i]<a[j])                add(b[i],a[j],1,0);        }        printf("%d\n",-MCMF(100001,100004));    }    return 0;}