POJ 3680 Intervals 离散 + 费用流

Intervals

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 6246		Accepted: 2542

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

题目大意：给你 n 个开区间，区间有权值，问如何选择区间使得每个点上覆盖的区间数不超过 k 个的前提下获得的最大权值和。

题目分析：做之前正好看过大白鼠的方法，所以过的没压力。首先，先对区间的端点进行离散化，之后对每个区间的左端点u和右端点v建边（u，v，1，-w），再对每个坐标点 i 建边（i，i + 1，k，0），设离散后最大的点的坐标为x，建立源汇点s = 0， t = x + 1，建边（x，t，1，0）。跑一次最小费用最大流，结果即花费的相反数。

代码如下：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define REP(i, n) for(int i = 0; i < n; ++i)
#define MS0(X) memset(X,  0, sizeof X)
#define MS1(X) memset(X, -1, sizeof X)
using namespace std;
const int maxE = 3000000;
const int maxN = 500;
const int maxM = 55;
const int oo = 0x3f3f3f3f;
struct Edge{
    int v, c, w, n;
};
Edge edge[maxE];
int adj[maxN], l;
int d[maxN], cur[maxN], Minflow;
int inq[maxN], Q[maxE], head, tail;
int cost, flow, s, t;
int n, m, cnt;
struct Node{
    int l, r, w;
}a[maxN];
int b[maxN];
void addedge(int u, int v, int c, int w){
    edge[l].v = v; edge[l].c = c; edge[l].w =  w; edge[l].n = adj[u]; adj[u] = l++;
    edge[l].v = u; edge[l].c = 0; edge[l].w = -w; edge[l].n = adj[v]; adj[v] = l++;
}
int SPFA(){
    memset(d, oo, sizeof d);
    memset(inq, 0, sizeof inq);
    head = tail = 0;
    d[s] = 0;
    Minflow = oo;
    cur[s] = -1;
    Q[tail++] = s;
    while(head != tail){
        int u =  Q[head++];
        inq[u] = 0;
        for(int i = adj[u]; ~i; i = edge[i].n){
            int v = edge[i].v;
            if(edge[i].c && d[v] > d[u] + edge[i].w){
                d[v] = d[u] + edge[i].w;
                cur[v] = i;
                Minflow = min(edge[i].c, Minflow);
                if(!inq[v]){
                    inq[v] = 1;
                    Q[tail++] = v;
                }
            }
        }
    }
    if(d[t] == oo) return 0;
    flow += Minflow;
    cost += Minflow * d[t];
    for(int i = cur[t]; ~i; i = cur[edge[i ^ 1].v]){
        edge[i].c -= Minflow;
        edge[i ^ 1].c += Minflow;
    }
    return 1;
}
int MCMF(){
    flow = cost = 0;
    while(SPFA());
    return cost;
}
void work(){
    MS1(adj);
    l = 0;
    scanf("%d%d", &n, &m);
    REP(i, n) scanf("%d%d%d", &a[i].l, &a[i].r, &a[i].w);
    cnt = 0;
    REP(i, n){
        b[cnt++] = a[i].l;
        b[cnt++] = a[i].r;
    }
    sort(b, b + cnt);
    int cnt1 = 0;
    REP(i, cnt) if(i && b[i] != b[cnt1]) b[++cnt1] = b[i];
    cnt = ++cnt1;
    //不会离散化就这么蠢蠢的离散好了QvQ
    REP(i, n) REP(j, cnt) if(a[i].l == b[j]){
        a[i].l = j;
        break;
    }
    REP(i, n) REP(j, cnt) if(a[i].r == b[j]){
        a[i].r = j;
        break;
    }
    s = 0; t = cnt;
    REP(i, n) addedge(a[i].l, a[i].r, 1, -a[i].w);
    REP(i, cnt) addedge(i, i + 1, m, 0);
    printf("%d
", -MCMF());
}
int main(){
    int T, cas;
    for(scanf("%d", &T), cas = 1; cas <= T; ++cas) work();
    return 0;
}

POJ 3680