题目链接：https://codeforces.com/contest/617
A - Elephant
事到如今这种三行签到是真的送的真诚。
B - Chocolate
题意：要把一块巧克力长条分成若干段，让每段都恰有1颗榛子，求方案数。
题解：每段被1包围的0，提供其数量+1的方案数。dp解法的话好像要分几种情况写的。
C - Watering Flowers
题意：有n<=2000个点，每个点表示一盆花。给花浇水，每盆花至少要1号喷头或者2号喷头其中之一喷到（指在圆内）。最小化喷头的圆的总面积。
题解：数据量小，直接暴力，枚举一个点是1号喷头的边界点，剩下的喷不到的就是2号喷头的。（这样会漏掉全部都由2号喷头来喷的情况！）
ll x[2005], y[2005];
void test_case() {
int n;
ll x1, y1, x2, y2;
cin >> n >> x1 >> y1 >> x2 >> y2;
for(int i = 1; i <= n; ++i)
cin >> x[i] >> y[i];
ll ans = LINF;
for(int i = 1; i <= n; ++i) {
ll R1 = (x[i] - x1) * (x[i] - x1) + (y[i] - y1) * (y[i] - y1);
ll maxR2 = 0;
for(int j = 1; j <= n; ++j) {
ll dis1 = (x[j] - x1) * (x[j] - x1) + (y[j] - y1) * (y[j] - y1);
if(dis1 > R1) {
ll dis2 = (x[j] - x2) * (x[j] - x2) + (y[j] - y2) * (y[j] - y2);
maxR2 = max(maxR2, dis2);
}
}
ans = min(ans, R1 + maxR2);
}
ll maxR2 = 0;
for(int j = 1; j <= n; ++j) {
ll dis2 = (x[j] - x2) * (x[j] - x2) + (y[j] - y2) * (y[j] - y2);
maxR2 = max(maxR2, dis2);
}
ans = min(ans, maxR2);
cout << ans << endl;
}
D
分类讨论进阶，我吐了。
void test_case() {
ll x1, y1, x2, y2, x3, y3;
cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
int ans = 3;
if(x1 == x2) {
if(x2 == x3)
ans = min(ans, 1);
else if(y3 >= max(y1, y2) || y3 <= min(y1, y2))
ans = min(ans, 2);
else
ans = min(ans, 3);
} else if(y1 == y2) {
if(y2 == y3)
ans = min(ans, 1);
else if(x3 >= max(x1, x2) || x3 <= min(x1, x2))
ans = min(ans, 2);
else
ans = min(ans, 3);
} else {
if(x1 == x3 || x2 == x3) {
if(y1 > y2)
swap(x1, x2), swap(y1, y2);
if(y3 <= max(y1, y2) && y3 >= min(y1, y2))
ans = min(ans, 2);
else if(x1 == x3 && y3 <= y1)
ans = min(ans, 2);
else if(x2 == x3 && y3 >= y2)
ans = min(ans, 2);
else
ans = min(ans, 3);
} else if(y1 == y3 || y2 == y3) {
if(x1 > x2)
swap(x1, x2), swap(y1, y2);
if(x3 <= max(x1, x2) && x3 >= min(x1, x2))
ans = min(ans, 2);
else if(y1 == y3 && x3 <= x1)
ans = min(ans, 2);
else if(y2 == y3 && x3 >= x2)
ans = min(ans, 2);
else
ans = min(ans, 3);
} else
ans = min(ans, 3);
}
cout << ans << endl;
}