POJ 3565 Ants (最小权匹配)

题意

给出一些蚂蚁的点，给出一些树的点，两两对应，使他们的连线不相交，输出一种方案。

思路

一开始没想到怎么用最小权匹配……后来发现是因为最小权匹配的方案一定不相交（三角形两边之和大于第三边）……还是too young too simple……没有融会贯通……

代码

[cpp]
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <cstring>
#define MID(x,y) ((x+y)/2)
#define MEM(a,b) memset(a,b,sizeof(a))
#define REP(i, begin, m) for (int i = begin; i < begin+m; i ++)
using namespace std;
const int MAXV = 205; //X or Y点集大小
const int oo = 0x3fffffff;

const double eps = 1e-8;
template <class weight_type>
struct MaximalMatchingOfWeightedBipartiteGraph{
weight_type w[MAXV][MAXV]; //权值
int sv, tv; //Perfect Matching, sv should equal to tv
bool S[MAXV], T[MAXV];
weight_type lx[MAXV], ly[MAXV]; //X、Y点集可行顶标
int left[MAXV];
weight_type slack[MAXV];
void init(int v){
sv = tv = v;
MEM(w, 0);
}
void add_uedge(int u, int v, weight_type _w){
w[u][v] = _w;
}
bool cross_path(int u){
S[u] = true;
for (int v = 1; v <= tv; v ++){
if(T[v]) continue;
weight_type t = lx[u] + ly[v] - w[u][v];
if (t < eps){
T[v] = true;
if (left[v] == 0 || cross_path(left[v])){
left[v] = u;
return true;
}
}
else{
slack[v] = min(slack[v], t);
}
}
return false;
}
weight_type solve(){
//Init
MEM(left, 0);
for (int i = 1; i <= sv; i ++){
lx[i] = 0;
ly[i] = 0;
for (int j = 1; j <= tv; j ++)
lx[i] = max(lx[i], w[i][j]);
}
//Main
for (int i = 1; i <= sv; i ++){
for (int j = 1; j <= tv; j ++) slack[j] = oo;
while(1){
MEM(S, false); MEM(T, false);
if (cross_path(i)){
break;
}
else{
weight_type d = oo;
for (int j = 1; j <= tv; j ++)
if (!T[j]) d = min(d, slack[j]);
for (int j = 1; j <= sv; j ++)
if (S[j]) lx[j] -= d;
for (int j = 1; j <= tv; j ++){
if (T[j]) ly[j] += d;
else slack[j] -= d;
//匈牙利树中T集点ly不变,S集点lx减小,更新slack值
}
}
}
}
weight_type res = 0;
for(int i = 1; i <= sv; i ++) res += lx[i];
for(int i = 1; i <= tv; i ++) res += ly[i];
return res;
}
};
MaximalMatchingOfWeightedBipartiteGraph <double> km;

struct Point{
double x, y;
}p[MAXV];
double dist(int i, int j){
return sqrt((p[i].x - p[j].x)*(p[i].x - p[j].x) + (p[i].y - p[j].y)*(p[i].y - p[j].y));
}

int main(){
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
int n;
while(scanf("%d", &n) != EOF){
REP(i, 1, n){
scanf("%lf %lf", &p[i+n].x, &p[i+n].y);
}
REP(i, 1, n){
scanf("%lf %lf", &p[i].x, &p[i].y);
}
km.init(n);
REP(i, 1, n){
REP(j, 1, n){
km.add_uedge(i, j, -dist(i, j+n));
}
}
//printf("%f\n", -km.solve());
km.solve();
REP(i, 1, n){
printf("%d\n", km.left[i]);
}
}
return 0;
}
[/cpp]