LightOJ 1219 Mafia(dfs—树上贪心)

1219 - Mafia

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Time Limit: 1 second(s)Memory Limit: 32 MB

Don Seliari gives his right hand Tommas Angelo (Tommy) an important job to guard his territory. The mafia territory consists ofn cities. And the cities are connected to each other in such a way that there is exactly one path from any city to another. Tommy has exactlyn mafia boys to guard the cities. Initially the mafia boys are resting randomly at then cities. Tommy wants every city to be guarded by the mafia boys. This is to be accomplished by a sequence of moves; each move consists of moving one mafia boy to the adjacent city. What is the minimum number of moves required so that every city is guarded?

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing an integern (1 ≤ n ≤ 10000).Each of the next n line contains at least three numbers which are:v the number of a city, followed by the number of mafia boys placed at cityv followed by a number d which is the number of cities adjacent ofv, followed by d numbers giving the adjacent cities ofv. Amongst the cities, one city is chosen as root, and for any city all its adjacent cities are listed except the one that connects it (through a path) to the root city.

Output

For each case, print the case number and the minimum number of moves required for the mafia boys to guard all cities.

Sample Input

Output for Sample Input

2

9

1 2 3 2 3 4

2 1 0

3 0 2 5 6

4 1 3 7 8 9

5 3 0

6 0 0

7 0 0

8 2 0

9 0 0

9

1 0 3 2 3 4

2 0 0

3 0 2 5 6

4 9 3 7 8 9

5 0 0

6 0 0

7 0 0

8 0 0

9 0 0

Case 1: 7

Case 2: 14

 

这道题也是磨了好几天，vector建的单向边各种炸，不多说了，贴一个邻接表的代码吧

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define maxn 10010int val[maxn],mark[maxn],ans;struct node{int u,v,next;}edge[maxn];int head[maxn],top; void init(){memset(head,-1,sizeof(head));top=0;memset(mark,0,sizeof(mark));} void add(int u,int v){edge[top].v=v;edge[top].next=head[u];head[u]=top++;}void dfs(int u,int pre){for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;dfs(v,u);}ans+=abs(val[u]-1);val[pre]+=(val[u]-1);}int main(){int t,n,m,u,v,k=1,i;scanf("%d",&t);while(t--){init();scanf("%d",&n);for(i=0;i<n;++i){scanf("%d",&u);scanf("%d%d",&val[u],&m);while(m--){scanf("%d",&v);mark[v]++; add(u,v);}}int root=-1;for(i=1;i<=n;++i){if(!mark[i])root=i;}ans=0;dfs(root,-1);printf("Case %d: %d\n",k++,ans);}return 0;}