Intervals
Description
You are given N weighted open intervals. The ith interval covers (ai, bi) 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 ai, bi, wi(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
題意：給出一些開區間，每個區間配一個權值。
問：在區間每個點不被重疊k次能取到的最大權值。
分析：區間內每個點不能被重疊k次。則每個點可以配一個容量為k的邊。設k為1，這樣一段區間A包含區間B，則B的權限比A的低，因為B要選取得看看A是否被選取，如果A被選取，則A的左端點已經滿了，所以B在權值小于A的話是不被選的。對其他重疊情況也是如此，這樣保證了每個區間之多被重疊k次。感覺這樣構圖之后就成了邊連通度的模型了。由于很多點可以被離散化，所以建成圖的點數就少了。最多也就401個點。這個經典的構圖太帥了，可惜不是自己想的。
代碼：
#include <stdio.h>
#include <stdlib.h>
#include <queue>
#define maxn 1010
#define inf (1 << 28)
using namespace std;
struct Edge
{
int u, v, c, f, w, next;
}e[210000];
int cnt;
int head[maxn], dis[maxn], visit[maxn], pos[maxn];
void addEdge(int u, int v, int c, int w)
{
e[cnt].u = u, e[cnt].v = v, e[cnt].c = c, e[cnt].f = 0;
e[cnt].w = w, e[cnt].next = head[u], head[u] = cnt++;
e[cnt].u = v, e[cnt].v = u, e[cnt].c = 0, e[cnt].f = 0;
e[cnt].w = -w, e[cnt].next = head[v], head[v] = cnt++;
}
void spfa(int s, int t, int ver)
{
int i, j, u, v;
for (i = 0; i <= ver; i++)
{
visit[i] = 0;
dis[i] = -1;
pos[i] = -1;
}
queue <int> que;
que.push(s);
dis[s] = 0;
visit[s] = 1;
while (!que.empty())
{
u = que.front();
que.pop();
visit[u] = 0;
for (i = head[u]; i + 1; i = e[i].next)
{
v = e[i].v;
if (e[i].c > e[i].f && dis[v] < dis[u] + e[i].w)
{
dis[v] = dis[u] + e[i].w;
pos[v] = i;
if (!visit[v])
{
visit[v] = 1;
que.push(v);
}
}
}
}
}
int CostFlow(int s, int t, int ver)
{
int i;
int cost = 0, flow = 0, minf;
while (1)
{
spfa(s, t, ver);
if (dis[t] == -1)
{
break;
}
minf = inf;
for (i = pos[t]; i != -1; i = pos[e[i].u])
{
minf = min(minf, e[i].c - e[i].f);
}
for (i = pos[t]; i != -1; i = pos[e[i].u])
{
e[i].f += minf, e[i^1].f -= minf;
}
flow += minf;
cost += minf * dis[t];
}
return cost;
}
int interval[maxn][3], lsh[maxn];
int cmp(const void * a, const void * b)
{
return *((int*)a) - *((int*)b);
}
int bsc(int l, int r, int x)
{
int mid;
while (l <= r)
{
mid = (l + r ) / 2;
if (lsh[mid] < x)
{
l = mid + 1;
}
else if (lsh[mid] > x)
{
r = mid - 1;
}
else
{
return mid;
}
}
}
int main()
{
int ca, n, k, i, j, m;
scanf("%d", &ca);
while (ca--)
{
scanf("%d%d", &n, &k);
for (i = 0, j = 0; i < n; i++)
{
scanf("%d%d%d", &interval[i][0], &interval[i][1], &interval[i][2]);
lsh[j++] = interval[i][0], lsh[j++] = interval[i][1];
}
qsort(lsh, j, sizeof(int), cmp);
m = j;
for (i = 0, j = 1; j < m; j++)
{
if (lsh[i] != lsh[j])
{
lsh[++i] = lsh[j];
}
}
m = i;
int s = 0, t = m + 1, ver = t + 1;
int u, v;
cnt = 0;
for (i = 0; i <= ver; i++) head[i] = -1;
for (i = 0; i <= m; i++)
{
addEdge(i, i + 1, k, 0);
}
for (i = 0; i < n; i++)
{
u = bsc(0, m, interval[i][0]) + 1;
v = bsc(0, m, interval[i][1]) + 1;
addEdge(u, v, 1, interval[i][2]);
}
int ans = CostFlow(s, t, ver);
printf("%d\n", ans);/**/
}
return 0;
}