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 (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

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

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;
}

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POJ 3680 Intervals（费用流）

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