poj 1034 The dog task

二分图的最大匹配。

建图的时候从主人路径上的前（n - 1）个点出发，枚举m个景点，能满足条件的连边，然后求最大匹配，输出路径就可以了。

/*
* Author: stormdpzh
* Created Time: 2012/9/11 20:14:45
* File Name: b.cpp
*/
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <list>
#include <algorithm>
#include <functional>

#define sz(v) ((int)(v).size())
#define rep(i, n) for(int i = 0; i < n; i++
#define repf(i, a, b) for(int i = a; i <= b; i++)
#define repd(i, a, b) for(int i = a; i >= b; i--)
#define out(n) printf("%d\n", n)
#define mset(a, b) memset(a, b, sizeof(a))
#define lint long long

using namespace std;

const int INF = 1 << 30;
const int MaxN = 105;
const double eps = 1e-8;

struct Node {
int x, y;
void read()
{
scanf("%d%d", &x, &y);
}
};
Node pnt[MaxN], dog[MaxN];
int n, m;
bool mp[MaxN][MaxN];
int x[MaxN], y[MaxN];
bool vis[MaxN];

int sgn(double d)
{
if(d > eps) return 1;
if(d < -eps) return -1;
return 0;
}

bool check(int id1, int id2)
{
int xx = pnt[id1 + 1].x - pnt[id1].x;
int yy = pnt[id1 + 1].y - pnt[id1].y;
double total = xx * xx + yy * yy;
int xx1 = dog[id2].x - pnt[id1].x, xx2 = pnt[id1 + 1].x - dog[id2].x;
int yy1 = dog[id2].y - pnt[id1].y, yy2 = pnt[id1 + 1].y - dog[id2].y;
int total1 = xx1 * xx1 + yy1 * yy1;
int total2 = xx2 * xx2 + yy2 * yy2;
if(sgn(sqrt((double)total2) + sqrt((double)total1) - 2 * sqrt((double)total)) <= 0) return true;
return false;
}

bool find(int u)
{
for(int i = 0; i < m; i++) {
if(!vis[i] && mp[u][i]) {
vis[i] = true;
if(y[i] == -1 || find(y[i])) {
x[u] = i;
y[i] = u;
return true;
}
}
}
return false;
}

int gao()
{
int match = 0;
memset(x, -1, sizeof(x));
memset(y, -1, sizeof(y));
for(int i = 0; i < n - 1; i++) {
if(x[i] == -1) {
memset(vis, false, sizeof(vis));
if(find(i)) match++;
}
}
return match;
}

int main()
{
while(2 == scanf("%d%d", &n, &m)) {
for(int i = 0; i < n; i++) pnt[i].read();
for(int i = 0; i < m; i++) dog[i].read();
mset(mp, false);
for(int i = 0; i < n - 1; i++) {
for(int j = 0; j < m; j++) {
if(check(i, j)) mp[i][j] = true;
}
}
int res = gao();
printf("%d\n", res + n);
for(int i = 0; i < n - 1; i++) {
if(x[i] != -1) printf("%d %d %d %d ", pnt[i].x, pnt[i].y, dog[x[i]].x, dog[x[i]].y);
else printf("%d %d ", pnt[i].x, pnt[i].y);
}
printf("%d %d\n", pnt[n - 1].x, pnt[n - 1].y);
}
return 0;
}