POJ3680_Intervals_离散化&&最小费用流

题意

思路

题目链接

AC代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>

using namespace std;

typedef pair<int, int> P;
struct edge
{
int to, cap, cost, rev;
edge(int a, int b, int c, int d)
:to(a), cap(b), cost(c), rev(d){}
};

const int maxn = 210;
const int maxv = maxn * 2;
const int  inf = 0x3f3f3f3f;

int cas;
int n, k;
int A[maxn], B[maxn], C[maxn];                                      //保存区间信息
vector<int> X;                                                      //用于离散化

//最小费用流零件
vector<edge> G[maxv];
int dis0[maxv], dis1[maxv];
int pv[maxv], pe[maxv];

//向图中添加边
void add(int from, int to, int cap, int cost)
{
G[from].push_back(edge(to, cap, cost, G[to].size()));
G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));
}

//迪杰斯特拉算法求最短路
void dijkstra(int s)
{
memset(dis1, inf, sizeof dis1);
dis1[s] =
4000
0;
priority_queue<P, vector<P>, greater<P> > qu;
qu.push(P(0, s));

while(qu.size())
{
P p = qu.top();qu.pop();
int v = p.second;
if(dis1[v] < p.first) continue;

for(int i= 0; i< G[v].size(); i++)
{
edge e = G[v][i];
if(e.cap > 0 && dis1[e.to] > dis1[v] + e.cost + dis0[v] - dis0[e.to])
{
dis1[e.to] = dis1[v] + e.cost + dis0[v] - dis0[e.to];
pv[e.to] = v;
pe[e.to] = i;
qu.push(P(dis1[e.to], e.to));
}
}
}
}

//最小费用流算法
int min_cost_flow(int s, int t, int f)
{
int res = 0;
memset(dis0, 0, sizeof dis0);

while(f > 0)
{
dijkstra(s);

if(dis1[t] == inf) return -1;

for(int i= s; i<= t; i++)
dis0[i] += dis1[i];

int d = f;
for(int v= t; v!= s; v= pv[v])
d = min(d, G[pv[v]][pe[v]].cap);

f -= d;
res += d * dis0[t];

for(int v= t; v!= s; v= pv[v])
{
edge &e = G[pv[v]][pe[v]];
e.cap -= d;
G[e.to][e.rev].cap += d;
}
}

return res;
}

int main()
{
cin >> cas;
while(cas--)
{
X.clear();
for(int i= 0; i< maxv; i++)
G[i].clear();

cin >> n >> k;

for(int i= 0; i< n; i++)
{
cin >> A[i] >> B[i] >> C[i];
X.push_back(A[i]);
X.push_back(B[i]);
}

//离散化
sort(X.begin(), X.end());
X.erase(unique(X.begin(), X.end()), X.end());

int s = 0, t = X.size() - 1;

for(int i= 0; i< t; i++)
add(i, i+1, inf, 0);

for(int i= 0; i< n; i++)
{
int a = lower_bound(X.begin(), X.end(), A[i]) - X.begin();
int b = lower_bound(X.begin(), X.end(), B[i]) - X.begin();
add(a, b, 1, -C[i]);
}

cout << -min_cost_flow(s, t, k) << endl;
}

return 0;
}