POJ 3660 Alice and Bob's Trip //2010 Asia Regional Harbin

Alice and Bob's Trip

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 617    Accepted Submission(s): 169


Problem Description
Alice and Bob are going on a trip. Alice is a lazy girl who wants to minimize the total travelling distance, while Bob as an active boy wants to maximize it. At the same time, they cannot let the value to be less than a given integer L since that will make them miss too much pleasure, and they cannot let the value to be greater than a given integer R since they don't want to get too exhausted.
The city they are visiting has n spots and the spots are connected by directed edges. The spots are connected in such a way that they form a tree and the root will always be at spot 0. They take turns to select which edge to go. Both of them choose optimally. Bob will go first.
 

Input
There are multiple test cases. For every test case, the first line has three integers, n, L and R (1<=n<=500000, 0<=L, R<=1000000000). The next n-1 lines each has three integers a, b and c, indicating that there is an edge going from spot a to spot b with length c (1<=c<=1000). The spots are labeled from 0 to n-1.
There is a blank line after each test case.
Proceed to the end of file.
 

Output
If the total distance is not within the range [L, R], print "Oh, my god!" on a single line. Otherwise, print the most value Bob can get.
 

Sample Input

3 2 4
0 1 1
0 2 5

7 2 8
0 1 1
0 2 1
1 3 1
1 4 10
2 5 1
2 6 5

7 4 8
0 1 1
0 2 1
1 3 1
1 4 2
2 5 1
2 6 5

4 2 6
0 1 1
1 2 1
1 3 5

 

Sample Output

Oh, my god!
2
6
2

 

Source
2010 Asia Regional Harbin
 

Recommend
lcy

好水的树状DP，竟然写完了，编译没错误，而且1Y

 

 

#include<cstdio>

#include<cstring>

const int V=500010;

const int inf=0x7fffffff;

struct EDGE

{

    int v,val,next;

}edge[V*2];

int head[V];

int n,l,r;

int e;

 

void addedge(int u,int v,int c)

{

    edge[e].v=v;

    edge[e].val=c;

    edge[e].next=head[u];

    head[u]=e++;

}

 

int dfs(int u,int sum,int mark)

{

    int v,w;

    if(mark==1)

    {

        int big=-1;

        for(int i=head[u];i!=-1;i=edge[i].next)

        {

            v=edge[i].v;

            w=edge[i].val;

            if(sum+w>r)  continue;

            if(head[v]==-1)

            {

                if(sum+w<l) continue;

                if(sum+w>big)  big=sum+w;

            }

            else

            {

                int t=dfs(v,sum+w,0);

                if(t==-1)  continue;

                if(t>big) big=t;

            }

        }

        if(big==-1)  return -1;

        else return big;

    }

    else

    {

        int sam=inf;

        for(int i=head[u];i!=-1;i=edge[i].next)

        {

            v=edge[i].v;

            w=edge[i].val;

            if(sum+w>r) continue;

            if(head[v]==-1)

            {

                if(sum+w<l) continue;

                if(sum+w<sam) sam=sum+w;

            }

            else

            {

                int t=dfs(v,sum+w,1);

                if(t==-1) continue;

                if(t<sam) sam=t;

            }

        }

        if(sam!=inf) return sam;

        else return -1;

    }

}

int main()

{

    while(scanf("%d%d%d",&n,&l,&r)!=EOF)

    {

        e=0;

        memset(head,-1,sizeof(head));

        for(int i=1;i<n;i++)

        {

            int a,b,c;

            scanf("%d%d%d",&a,&b,&c);

            addedge(a,b,c);

        }

        int t=dfs(0,0,1);

        if(t<=r&&t>=l)  printf("%d/n",t);

        else printf("Oh, my god!/n");

    }

    return 0;

}