标签:memory tac print code 联系人 problem and max ase
你能帮Wiskey计算出至少要通知多少人。至少得花多少电话费就能让全部人都被通知到吗?
12 16 2 2 2 2 2 2 2 2 2 2 2 2 1 3 3 2 2 1 3 4 2 4 3 5 5 4 4 6 6 4 7 4 7 12 7 8 8 7 8 9 10 9 11 10
3 6
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;
const int MAX=1000+10;
int n,m,size,top,index;
int head[MAX],val[MAX],dfn[MAX],low[MAX];
int mark[MAX],stack[MAX];
struct Edge{
int v,next;
Edge(){}
Edge(int V,int NEXT):v(V),next(NEXT){}
}edge[MAX*2];
void Init(int num){
for(int i=0;i<=num;++i)head[i]=-1,mark[i]=0;
size=top=index=0;
}
void InsertEdge(int u,int v){
edge[size]=Edge(v,head[u]);
head[u]=size++;
}
void tarjan(int u){
if(mark[u])return;
dfn[u]=low[u]=++index;
stack[++top]=u;
mark[u]=1;
for(int i=head[u];i != -1;i=edge[i].next){
int v=edge[i].v;
tarjan(v);
if(mark[v] == 1)low[u]=min(low[u],low[v]);
}
if(dfn[u] == low[u]){
while(stack[top] != u){
mark[stack[top]]=-1;
val[u]=min(val[u],val[stack[top]]);
low[stack[top--]]=low[u];
}
mark[u]=-1;
--top;
}
}
int main(){
int u,v;
while(~scanf("%d%d",&n,&m)){
Init(n);
for(int i=1;i<=n;++i)scanf("%d",val+i);
for(int i=0;i<m;++i){
scanf("%d%d",&u,&v);
InsertEdge(u,v);
}
for(int i=1;i<=n;++i){
if(mark[i])continue;
tarjan(i);
}
for(int i=0;i<=n;++i)mark[i]=0;
for(int i=1;i<=n;++i){
for(int j=head[i];j != -1;j=edge[j].next){
v=edge[j].v;
if(low[i] == low[v])continue;
mark[low[v]]=1;
}
}
int sum=0,ans=0;
for(int i=1;i<=n;++i){
if(!mark[low[i]] && dfn[i] == low[i])++ans,sum+=val[i];
mark[low[i]]=1;
}
printf("%d %d\n",ans,sum);
}
return 0;
}
标签:memory tac print code 联系人 problem and max ase
原文地址:http://www.cnblogs.com/cxchanpin/p/6849919.html