POJ 3680 Intervals(费用流)

Intervals

Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 5762 Accepted: 2288

Description

You are given N weighted open intervals. The ith interval covers (aibi) and weighs wi. 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 aibiwi(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

4

3 1
1 2 2
2 3 4
3 4 8

3 1
1 3 2
2 3 4
3 4 8

3 1
1 100000 100000
1 2 3
100 200 300

3 2
1 100000 100000
1 150 301
100 200 300

Sample Output

14
12
100000
100301

Source

POJ Founder Monthly Contest – 2008.07.27, windy7926778 

 

一条线段看成两个点。

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <string>
#include <queue>
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-1
void 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;
}

 

 

    原文作者:kuangbin
    原文地址: https://www.cnblogs.com/kuangbin/p/3236051.html
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