本來以爲是一道凸包題目,結果最後看了位大佬的題解才發現是圖論的算法。
大佬博客鏈接
首先對守衛熊的m個點兩兩一枚舉,對於每一次枚舉的兩個點a, b,去測試所有的n個村莊是否全在這次枚舉線段的一側,如果所有的點都在ab的左邊就把m個點中的ab連接一條有向邊,如果都在右邊,就對ba連一條邊,如果全在這條線段就把ab連接一條雙向邊,除此以外都不連。這樣處理後對m個點所建立的圖套模板跑一個floyd最小環就可以得到答案了。
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
#define Max_N 500+10
#define INF 0x3f3f3f3f
int n, m;
struct point{
int x, y;
};
point p_n[Max_N];
point p_m[Max_N];
int G[Max_N][Max_N];
int multi(point p1, point p2, point p0)
{
int ans = (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
if (ans > 0) return 1;
if (ans == 0) return 0;
if (ans < 0) return 2;
}
int main()
{
while (scanf("%d", &n) != EOF) {
memset(G, INF, sizeof(G));
for (int i = 1; i <= n; i++)
scanf("%d%d", &p_n[i].x, &p_n[i].y);
scanf("%d", &m);
for (int i = 1; i <= m; i++)
scanf("%d%d", &p_m[i].x, &p_m[i].y);
int flag = 4;
int flag1 = 0;
for (int i = 1; i <= m; i++)
for (int j = i+1; j <= m; j++) {
flag1 = 0;
flag = 4;
for (int k = 1; k <= n; k++) {
int mu = multi(p_n[k], p_m[i], p_m[j]);
if (mu == 0) {
if (min(p_m[i].x, p_m[j].x) <= p_n[k].x && max(p_m[i].x, p_m[j].x) >= p_n[k].x) continue;
else {
flag1 = 1;
break;
}
}
if (flag == 4) {
flag = mu;
}
else {
if (flag != mu) {
flag1 = 1;
break;
}
}
}
if (!flag1) {
if (flag == 1 || flag == 4) G[j][i] = 1;
if (flag == 2 || flag == 4) G[i][j] = 1;
}
}
int ans = INF;
for(int k=1;k<=m;k++)
{
for(int i=1;i<=m;i++)
{
if(G[i][k]==INF)
continue;
for(int j=1;j<=m;j++)
G[i][j] = min(G[i][j], G[i][k]+G[k][j]);
}
}
for(int i=1;i<=m;i++)
ans = min(ans, G[i][i]);
if(ans > m)
printf("ToT\n");
else
printf("%d\n", m-ans);
}
}