hdu 5638 Toposort (拓扑排序+线段树）

Toposort

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 499    Accepted Submission(s): 203

Problem Description

There is a directed acyclic graph with  n vertices and m edges. You are allowed to delete exact k edges in such way that the lexicographically minimal topological sort of the graph is minimum possible.

 

 

Input

There are multiple test cases. The first line of input contains an integer  T indicating the number of test cases. For each test case:

The first line contains three integers n, m and k (1≤n≤100000,0≤k≤m≤200000) -- the number of vertices, the number of edges and the number of edges to delete.

For the next m lines, each line contains two integers ui and vi, which means there is a directed edge from ui to vi (1≤ui,vi≤n).

You can assume the graph is always a dag. The sum of values of n in all test cases doesn't exceed 106. The sum of values of m in all test cases doesn't exceed 2×106.

 

 

Output

For each test case, output an integer  S=(∑i=1ni⋅pi) mod (109+7), where p1,p2,...,pn is the lexicographically minimal topological sort of the graph.

 

 

Sample Input

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

 

 

Sample Output

30 27 30

 

思路：

和上道hdu 5195差不多，改点东西就行了。。之前疯狂超时，找了一个小时找不到哪里出了问题，最后问了下徐巨，发现vector忘了清空，清空下就好了。

 

实现代码：

#include
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid ll m = (l + r) >> 1
const int M = 2e5+10;
const int inf = 0x3f3f3f3f;
const int md = 1e9+7;
ll n,m,cnt,key;
vectorg[M];
ll sum[M<<2],du[M];

void pushup(ll rt){
    sum[rt] = min(sum[rt<<1],sum[rt<<1|1]);
}

void build(ll l,ll r,ll rt){
    if(l == r){
        sum[rt] = du[l];
        return;
    }
    mid;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(ll p,ll l,ll r,ll rt){
    if(l == r){
        sum[rt]--;
        return ;
    }
    mid;
    if(p <= m) update(p,lson);
    else  update(p,rson);
    pushup(rt);
}

void query(ll l,ll r,ll rt){
    if(l == r){
        cnt -= sum[rt];
        key = l;
        sum[rt] = inf;
        return ;
    }
    mid;
    if(sum[rt<<1] <= cnt) query(lson);
    else query(rson);
    pushup(rt);
}

int main(){
    ll u,v;
    ll t;
    scanf("%lld",&t);
    while(t--){
        scanf("%lld%lld%lld",&n,&m,&cnt);
        for(ll i = 1;i <= m;i ++){
            scanf("%lld%lld",&u,&v);
            g[u].push_back(v);
            du[v]++;
        }
        build(1,n,1);
        ll ans = 0;
        for(ll i = 1;i <= n;i ++){
            query(1,n,1);
            ans=(ans+key*i)%md;
            for(ll j = 0;j < g[key].size();j++){
                ll x = g[key][j];
                update(x,1,n,1);
            }
        }
        printf("%lld\n",ans);
        memset(du,0,sizeof(du));
        memset(sum,inf,sizeof(sum));
        for(ll i=0;i<=n;i++)
            g[i].clear();
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/kls123/p/8835466.html