蚂蚁
分析
见《进阶指南》第432
页。
实现
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#define inf 0x3f3f3f3f
using namespace std;
const int N = 110, M = 20410;
struct edge {
int to, next, c;
double w;
};
edge e[M];
int idx, head[2 * N];
struct node {
int x, y;
};
int n, s, t, max_flow, res[N];
double min_cost = 0;
node w[N], b[N];
double dis[2 * N];
int flow[2 * N], eid[2 * N];
bool mark[2 * N];
void add_edge (int u, int v, int c, double w) {
e[idx].c = c;
e[idx].w = w;
e[idx].to = v;
e[idx].next = head[u];
head[u] = idx ++;
e[idx].c = 0;
e[idx].w = -w;
e[idx].to = u;
e[idx].next = head[v];
head[v] = idx ++;
}
double get_dis (const node&a, const node& b) {
double dx = a.x - b.x, dy = a.y - b.y;
return sqrt(dx * dx + dy * dy);
}
bool spfa () {
memset(mark, 0, sizeof(mark));
for (int i = 1; i <= t; ++ i) dis[i] = 1e12;
dis[s] = 0;
flow[s] = inf;
queue<int> q;
q.push(s);
mark[s] = true;
while (!q.empty()) {
int cur = q.front();
q.pop();
mark[cur] = false;
for (int i = head[cur]; i != -1; i = e[i].next) {
int to = e[i].to, c = e[i].c;
double w = e[i].w;
if (c == 0) continue;
if (dis[cur] + w < dis[to]) {
dis[to] = dis[cur] + w;
flow[to] = min(flow[cur], c);
eid[to] = i;
if (mark[to] == false) {
q.push(to);
mark[to] = true;
}
}
}
}
return dis[t] != 1e12;
}
void update () {
int cur = t;
while (cur != s) {
int i = eid[cur];
e[i].c -= flow[t];
e[i ^ 1].c += flow[t];
cur = e[i ^ 1].to;
}
max_flow += flow[t];
min_cost += dis[t] * flow[t];
}
int main () {
memset(head, -1, sizeof(head));
cin >> n;
for (int i = 1; i <= n; ++ i) cin >> b[i].x >> b[i].y;
for (int i = 1; i <= n; ++ i) cin >> w[i].x >> w[i].y;
s = 0, t = 2 * n + 1;
for (int i = 1; i <= n; ++ i) {
add_edge(s, i, 1, 0);
add_edge(i + n, t, 1, 0);
}
for (int i = 1; i <= n; ++ i)
for (int j = 1; j <= n; ++ j)
add_edge(i, j + n, 1, get_dis(w[i], b[j]));
while (spfa() == true) update();
for (int i = 1; i <= n; ++ i) {
for (int j = head[i]; j != -1; j = e[j].next) {
int to = e[j].to, c = e[j].c;
if (c == 0) {
res[to - n] = i;
break;
}
}
}
for (int i = 1; i <= n; ++ i)
cout << res[i] << endl;
return 0;
}