Blog of RuSun

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

P1262 间谍网络

P1262 间谍网络

从接受贿赂的间谍缩点,如果有点没有遍历到输出 NO。

一个双连通分量的最小代价为其中一个点的最小代价,选择没有入度的点显然时最优的。

查看代码
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#include <iostream>
#include <cstdio>
#include <stack>
#include <climits>
#define inf INT_MAX
using namespace std;
const int N = 3e3 + 10, M = 8e3 + 10;
int n, m, w[N], money[N];
bool vis[N], din[N];
int idx, hd[N], nxt[M], edg[M];
int cnt, stmp, dfn[N], low[N], id[N];
stack<int> stk;
void tarjan(int x)
{
dfn[x] = low[x] = ++stmp;
stk.push(x);
vis[x] = true;
for (int i = hd[x]; ~i; i = nxt[i])
if (!dfn[edg[i]])
{
tarjan(edg[i]);
low[x] = min(low[x], low[edg[i]]);
}
else if (vis[edg[i]])
low[x] = min(low[x], dfn[edg[i]]);
if (dfn[x] == low[x])
{
int y;
cnt++;
do
{
y = stk.top();
stk.pop();
vis[y] = false;
id[y] = cnt;
} while (x != y);
}
}
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; i <= n; i++)
w[i] = money[i] = inf;
int q;
scanf("%d", &q);
for (int i = 1, a, b; i <= q; i++)
{
scanf("%d%d", &a, &b);
money[a] = b;
}
scanf("%d", &m);
for (int i = 1, a, b; i <= m; i++)
{
scanf("%d%d", &a, &b);
add(a, b);
}
for (int i = 1; i <= n; i++)
if (!dfn[i] && money[i] != inf)
tarjan(i);
for (int i = 1; i <= n; i++)
if (!dfn[i])
{
printf("NO\n%d", i);
return 0;
}
puts("YES");
for (int i = 1; i <= n; i++)
{
w[id[i]] = min(w[id[i]], money[i]);
for (int j = hd[i]; ~j; j = nxt[j])
if (id[i] != id[edg[j]])
din[id[edg[j]]] = true;
}
int res = 0;
for (int i = 1; i <= cnt; i++)
if (!din[i])
res += w[i];
printf("%d", res);
return 0;
}