感觉这题最大的难点是发现它的解法是二分图最大匹配
主人的路线是固定的,对于每一段的路线,我们可以枚举小狗想去的景点,如果时间够,我们就将这段路线的起点和小狗想去的点连起来
这样就形成了一个二分图,因为dog每次与主人相遇之前最多只去一个景点,所以这道题就转化成了二分图最大匹配
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
double X1[110], Y1[110], X2[110], Y2[110];
double calc(double a, double b, double c, double d) {
return sqrt((c-a) * (c-a) + (d-b) * (d-b));
}
struct zzz {
int t, nex;
}e[10010 << 1]; int head[110], tot;
void add(int x, int y) {
e[++tot].t = y;
e[tot].nex = head[x];
head[x] = tot;
}
int lin[110], ans; bool vis[110];
bool find(int x) {
for(int i = head[x]; i; i = e[i].nex) {
if(!vis[e[i].t]) {
vis[e[i].t] = 1;
if(!lin[e[i].t] || find(lin[e[i].t])) return lin[e[i].t] = x, 1;
}
}
return 0;
}
int step[110];
int main() {
int n = read(), m = read();
for(int i = 1; i <= n; ++i) X1[i] = read(), Y1[i] = read();
for(int i = 1; i <= m; ++i) X2[i] = read(), Y2[i] = read();
for(int i = 1; i <= n-1; ++i) {
double x = calc(X1[i], Y1[i], X1[i+1], Y1[i+1]);
for(int j = 1; j <= m; ++j) {
if((calc(X1[i], Y1[i], X2[j], Y2[j]) + calc(X2[j], Y2[j], X1[i+1], Y1[i+1])) <= x * 2)
add(i, j);
}
}
for(int i = 1; i < n; ++i) {
memset(vis, 0, sizeof(vis));
ans += find(i);
}
printf("%d
", ans+n);
for(int i = 1; i <= m; ++i) step[lin[i]] = i;
for(int i = 1; i <= n; ++i) {
printf("%.0lf %.0lf ", X1[i], Y1[i]);
if(step[i])
printf("%.0lf %.0lf ", X2[step[i]], Y2[step[i]]);
}
return 0;
}