UVA 1411 - Ants(二分图完美匹配)

UVA 1411 - Ants

题目链接

题意:给定一些黑点白点,要求一个黑点连接一个白点,而且全部线段都不相交

思路:二分图完美匹配,权值存负的欧几里得距离,这种话,相交肯定比不相交权值小,所以做一次完美匹配就能够了

代码:

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std; const int MAXNODE = 105; typedef double Type;
const Type INF = 0x3f3f3f3f; struct KM {
int n;
Type g[MAXNODE][MAXNODE];
Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];
int left[MAXNODE];
bool S[MAXNODE], T[MAXNODE]; void init(int n) {
this->n = n;
} void add_Edge(int u, int v, Type val) {
g[u][v] = val;
} bool dfs(int i) {
S[i] = true;
for (int j = 0; j < n; j++) {
if (T[j]) continue;
double tmp = Lx[i] + Ly[j] - g[i][j];
if (fabs(tmp) < 1e-9) {
T[j] = true;
if (left[j] == -1 || dfs(left[j])) {
left[j] = i;
return true;
}
} else slack[j] = min(slack[j], tmp);
}
return false;
} void update() {
double a = INF;
for (int i = 0; i < n; i++)
if (!T[i]) a = min(a, slack[i]);
for (int i = 0; i < n; i++) {
if (S[i]) Lx[i] -= a;
if (T[i]) Ly[i] += a;
}
} void km() {
for (int i = 0; i < n; i++) {
left[i] = -1;
Lx[i] = -INF; Ly[i] = 0;
for (int j = 0; j < n; j++)
Lx[i] = max(Lx[i], g[i][j]);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) slack[j] = INF;
while (1) {
for (int j = 0; j < n; j++) S[j] = T[j] = false;
if (dfs(i)) break;
else update();
}
}
}
} gao; const int N = 105;
int n;
struct Point {
double x, y;
void read() {
scanf("%lf%lf", &x, &y);
}
} a[N], b[N]; double dis(Point a, Point b) {
double dx = a.x - b.x;
double dy = a.y - b.y;
return sqrt(dx * dx + dy * dy);
} int main() {
int bo = 0;
while (~scanf("%d", &n)) {
if (bo) printf("\n");
else bo = 1;
gao.init(n);
for (int i = 0; i < n; i++)
a[i].read();
for (int i = 0; i < n; i++)
b[i].read();
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
gao.add_Edge(i, j, -dis(b[i], a[j]));
gao.km();
for (int i = 0; i < n; i++)
printf("%d\n", gao.left[i] + 1);
}
return 0;
}
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