Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P2458 [SDOI2006]保安站岗

P2458 [SDOI2006]保安站岗

注意一个坑点:题目没有说根节点为 $1$ 。

查看代码
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#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1510, M = 3e3 + 10, inf = 15e6;
bool notrt[N];
int n, rt, s[N], f[N][3];
int idx, hd[N], nxt[M], edg[M];
int dfs(int, int);
int cal(int x, bool op)
{
if (op)
{
int res = s[x];
for (int i = hd[x]; ~i; i = nxt[i])
res += dfs(edg[i], 0);
return res;
}
else
{
int res = 0;
for (int i = hd[x]; ~i; i = nxt[i])
res += min(dfs(edg[i], 1), dfs(edg[i], 2));
return res;
}
}
int dfs(int x, int op)
{
if (~f[x][op])
return f[x][op];
if (op == 0)
f[x][op] = min(cal(x, 0), cal(x, 1));
else if (op == 1)
f[x][op] = cal(x, 1);
else if (op == 2)
{
int s = 0;
for (int i = hd[x]; ~i; i = nxt[i])
s += min(dfs(edg[i], 1), dfs(edg[i], 2));
f[x][op] = inf;
for (int i = hd[x]; ~i; i = nxt[i])
f[x][op] = min(f[x][op], s - min(dfs(edg[i], 1), dfs(edg[i], 2)) + dfs(edg[i], 1));
}
return f[x][op];
}
void add(int a, int b)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1, a, m; i <= n; i++)
{
scanf("%d", &a);
scanf("%d%d", &s[a], &m);
for (int b; m; m--)
{
scanf("%d", &b);
add(a, b);
notrt[b] = true;
}
}
for (int i = 1; i <= n; i++)
if (!notrt[i])
rt = i;
memset(f, -1, sizeof f);
printf("%d", min(dfs(rt, 1), dfs(rt, 2)));
return 0;
}