BZOJ4813或洛谷3698 [CQOI2017]小Q的棋盘
BZOJ原题链接
洛谷原题链接
贪心或树形\(DP\)都可做,但显然\(DP\)式子不好推(因为我太菜了),所以我选择贪心。
很显然从根出发主干走最长链是最优的,而剩下的点每个都需要走两步,所以用除去走最长链的步数的剩余步数除\(2\)(下取整)就是剩余能走的点数。
#include<cstdio>
using namespace std;
const int N = 110;
const int M = N << 1;
int fi[N], di[M], ne[M], de[N], l, ma_de, ma_id;
inline int re()
{
int x = 0;
char c = getchar();
bool p = 0;
for (; c < '0' || c > '9'; c = getchar())
p |= c == '-';
for (; c >= '0' && c <= '9'; c = getchar())
x = x * 10 + c - '0';
return p ? -x : x;
}
inline void add(int x, int y)
{
di[++l] = y;
ne[l] = fi[x];
fi[x] = l;
}
inline int minn(int x, int y)
{
return x < y ? x : y;
}
void dfs(int x)
{
int i, y;
if (ma_de < de[x])
{
ma_de = de[x];
ma_id = x;
}
for (i = fi[x]; i; i = ne[i])
if (!de[y = di[i]])
{
de[y] = de[x] + 1;
dfs(y);
}
}
int main()
{
int i, n, m, x, y;
n = re();
m = re();
for (i = 1; i < n; i++)
{
x = re() + 1;
y = re() + 1;
add(x, y);
add(y, x);
}
de[1] = 1;
dfs(1);
if (m <= ma_de - 1)
{
printf("%d", m + 1);
return 0;
}
m -= ma_de - 1;
printf("%d", ma_de + minn(m >> 1, n - ma_de));
return 0;
}
posted on 2018-11-01 14:57 Iowa_Battleship 阅读(113) 评论(0) 编辑 收藏 举报