zoj3527

Shinryaku! Kero Musume

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Time Limit: 4 Seconds      Memory Limit: 65536 KB

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2300 years ago, Moriya Suwako was defeated by Yasaka Kanako in the Great Suwa War. As the goddess of mountains in Gensokyo, she was planning to make a comeback and regain faith among humans.

To achieve her great plan, she decides to build shrines in human villages. Of course, each village can bulid at most one shrine. If she builds a shrine in the i-th village, she can get F_ifaith points.

Because of geological differences and the Quantum Entanglement theory, each village has one and only one "entangled" village P_i (it is a kind of one-way relationship). If Suwakobuilds shrines both in the i-th village and the P_i-th village, the faith she get from the i-th village will changes by G_i points (it can result to a negative value of faith points). If she only builds a shrine in the P_i-th village, but not in the i-th village, the faith she get will not changes.

Now, please help Suwako to find out the maximal faith points she can get.

Input

There are multiple test cases. For each test case:

The first line contains an integer N (2 <= N <= 100000) indicates the number of villages in Gensokyo. Then followed by N lines, each line contains three integers F_i (0 <= F_i <= 100000) G_i (-100000 <= G_i <= 100000) P_i (1 <= P_i <= N and P_i will not point to the i-th village itself).

Output

For each test case, output the maximal faith points that Suwako can get.

Sample Input

23 -1 22 -2 143 -2 24 -3 32 -1 15 -2 2

Sample Output

39

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Author: JIANG, Kai

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;typedef long long ll;const int MAXN=100000+100;int n,m,edge_cnt;int a[MAXN],g[MAXN],in[MAXN],head[MAXN],vis[MAXN];ll dp[MAXN][2],dp1[MAXN][2];struct Edge{    int v;    int next;}edge[MAXN];void init(){    edge_cnt=0;    memset(in,0,sizeof(in));    memset(dp,0,sizeof(dp));    memset(vis,0,sizeof(vis));    memset(head,-1,sizeof(head));}void addedge(int u,int v){    edge[edge_cnt].v=v;    edge[edge_cnt].next=head[u];    head[u]=edge_cnt++;}ll dfs(int st,int cur,ll dp[][2],int flag){    int u=cur,v=edge[head[u]].v;    while(v!=st)    {        dp[v][0]+=max(dp[u][0],dp[u][1]);        dp[v][1]+=max(dp[u][0],dp[u][1]+g[u]);        u=v;        v=edge[head[u]].v;    }    if(flag)        dp[u][1]+=g[u];    return max(dp[u][0],dp[u][1]);}ll get_ans(int u){    int v=edge[head[u]].v;    dp1[v][0]+=dp[u][0];    dp1[v][1]+=dp[u][0];// u不取    ll ans1=dfs(u,v,dp1,0);    dp[v][0]+=dp[u][1];    dp[v][1]+=dp[u][1]+g[u];    ll ans2=dfs(u,v,dp,1);    return max(ans1,ans2);}int main(){    //freopen("text.txt","r",stdin);    while(~scanf("%d",&n))    {        init();        for(int i=1;i<=n;i++)        {            int j;            scanf("%d%d%d",&a[i],&g[i],&j);            addedge(i,j); in[j]++;        }        queue<int>Q;        for(int i=1;i<=n;i++)        {            dp[i][1]=a[i];            if(!in[i])                Q.push(i);        }        while(!Q.empty())        {            int u=Q.front(); Q.pop();            vis[u]=1;            int v=edge[head[u]].v;            dp[v][0]+=max(dp[u][0],dp[u][1]);            dp[v][1]+=max(dp[u][1]+g[u],dp[u][0]);            if(!(--in[v]))                Q.push(v);        }        memcpy(dp1,dp,sizeof(dp));        ll sum=0;        for(int i=1;i<=n;i++)            if(!vis[i])            {                sum+=get_ans(i);                vis[i]=1;                for(int j=edge[head[i]].v;j!=i;j=edge[head[j]].v)                    vis[j]=1;            }        printf("%lld\n",sum);    }    return 0;}