贿赂所有能贿赂的,如果还有人不被访问则显然是NO。
否则,必定为YES。强联通分量缩成一个DAG,若某点的入度为零,则答案要算上它的。
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
struct Edge{
int fro, too, nxt;
}edge[8005];
int n, p, uu, vv, scc, w[3005], cnt, hea[3005], dfn[3005], loo[3005], ind, sta[3005], din, r, ans;
int minn[3005], bel[3005], dge[3005];
bool ins[3005];
void add_edge(int fro, int too){
edge[++cnt].nxt = hea[fro];
edge[cnt].fro = fro;
edge[cnt].too = too;
hea[fro] = cnt;
}
void dfs(int u){
ins[u] = true;
for(int i=hea[u]; i; i=edge[i].nxt){
int t=edge[i].too;
if(!ins[t]) dfs(t);
}
}
void tarjan(int u){
sta[++din] = u;
ins[u] = true;
dfn[u] = loo[u] = ++ind;
for(int i=hea[u]; i; i=edge[i].nxt){
int t=edge[i].too;
if(!dfn[t]){
tarjan(t);
loo[u] = min(loo[u], loo[t]);
}
else if(ins[t]) loo[u] = min(loo[u], dfn[t]);
}
int j;
if(dfn[u]==loo[u]){
scc++;
do{
j = sta[din--];
bel[j] = scc;
ins[j] = false;
minn[scc] = min(minn[scc], w[j]);
}while(dfn[j]!=dfn[u]);
}
}
int main(){
cin>>n>>p;
memset(w, 0x3f, sizeof(w));
memset(minn, 0x3f, sizeof(minn));
for(int i=1; i<=p; i++) scanf("%d", &uu), scanf("%d", &w[uu]);
cin>>r;
for(int i=1; i<=r; i++){
scanf("%d %d", &uu, &vv);
add_edge(uu, vv);
}
for(int i=1; i<=n; i++)
if(w[i]!=0x3f3f3f3f && !ins[i])
dfs(i);
for(int i=1; i<=n; i++)
if(!ins[i]){
cout<<"NO
"<<i;
return 0;
}
memset(ins, 0, sizeof(ins));
for(int i=1; i<=n; i++)
if(!dfn[i])
tarjan(i);
for(int i=1; i<=cnt; i++)
if(bel[edge[i].fro]!=bel[edge[i].too])
dge[bel[edge[i].too]]++;
for(int i=1; i<=scc; i++){
if(!dge[i])
ans += minn[i];
}
cout<<"YES
"<<ans;
return 0;
}