luogu P1262 间谍网络 题解
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题目链接:https://www.luogu.org/problemnew/show/P1262
注意:
1.缩点时计算出入度是在缩完点的图上用color计算。不要在原来的点上计算。
2.枚举出入度时是在缩完点的图上计算。枚举范围到num。
#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 10000 + 10;
struct edge{
int from, next, to, len;
}e[maxn<<2];
int head[maxn], cnt;
int dfn[maxn], low[maxn], tim, color[maxn], num, rudu[maxn];
stack<int> s;
bool vis[maxn], okbuy[maxn], flag[maxn];
int n, p, m, money[maxn], minpay[maxn], ans;
void add(int u, int v)
{
e[++cnt].from = u;
e[cnt].next = head[u];
e[cnt].to = v;
head[u] = cnt;
}
void tarjan(int x)
{
dfn[x] = low[x] = ++tim;
s.push(x); vis[x] = 1;
for(int i = head[x]; i != -1; i = e[i].next)
{
int v = e[i].to;
if(!dfn[v])
{
tarjan(v);
low[x] = min(low[x], low[v]);
}
else if(vis[v])
{
low[x] = min(low[x], low[v]);
}
}
if(low[x] == dfn[x])
{
color[x] = ++num;
vis[x] = 0;
minpay[num] = min(minpay[num], money[x]);
while(s.top() != x)
{
color[s.top()] = num;
vis[s.top()] = 0;
minpay[num] = min(minpay[num], money[s.top()]);
s.pop();
}
s.pop();
}
}
int main()
{
memset(head, -1, sizeof(head));
scanf("%d%d",&n,&p);
for(int i = 1; i <= n; i++) minpay[i] = 0x7fffffff, money[i] = 0x7fffffff;
for(int i = 1; i <= p; i++)
{
int u, val;
scanf("%d%d",&u,&val);
okbuy[u] = 1;
money[u] = val;
}
scanf("%d",&m);
for(int i = 1; i <= m; i++)
{
int u, v;
scanf("%d%d",&u,&v);
add(u,v);
}
for(int i = 1; i <= n; i++)
if(!dfn[i] && okbuy[i] == 1) tarjan(i);
for(int i = 1; i <= n; i++)
if(!dfn[i])
{
printf("NO
%d",i);
return 0;
}
for(int i = 1; i <= n; i++)
for(int j = head[i]; j != -1; j = e[j].next)
{
int v = e[j].to;
if(color[v] != color[i])
{
rudu[color[v]]++;
}
}
for(int i = 1; i <= num; i++)
{
if(rudu[i] == 0)
ans += minpay[i];
}
printf("YES
%d
",ans);
return 0;
}
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