Go Deeper HDU - 3715(2 - sat 水题 妈的 智障)
Go Deeper
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3435 Accepted Submission(s): 1125
go(int dep, int n, int m)
begin
output the value of dep.
if dep < m and x[a[dep]] + x[b[dep]] != c[dep] then go(dep + 1, n, m)
end
In this code n is an integer. a, b, c and x are 4 arrays of integers. The index of array always starts from 0. Array a and b consist of non-negative integers smaller than n. Array x consists of only 0 and 1. Array c consists of only 0, 1 and 2. The lengths of array a, b and c are m while the length of array x is n. Given the elements of array a, b, and c, when we call the procedure go(0, n, m) what is the maximal possible value the procedure may output?
2 1
0 1 0
2 1
0 0 0
2 2
0 1 0
1 1 2
1
2
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define pd(a) printf("%d\n", a);
#define plld(a) printf("%lld\n", a);
#define pc(a) printf("%c\n", a);
#define ps(a) printf("%s\n", a);
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _ ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1e5 + , INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
int n, m;
int a[maxn], b[maxn], c[maxn];
vector<int> G[maxn];
int sccno[maxn], low[maxn], vis[maxn], scc_clock, scc_cnt;
stack<int> S;
void init()
{
for(int i = ; i < maxn; i++) G[i].clear();
mem(sccno, );
mem(low, );
mem(vis, );
scc_clock = scc_cnt = ;
} void dfs(int u)
{
low[u] = vis[u] = ++scc_clock;
S.push(u);
for(int i = ; i < G[u].size(); i++)
{
int v = G[u][i];
if(!vis[v])
{
dfs(v);
low[u] = min(low[u], low[v]);
}
else if(!sccno[v])
low[u] = min(low[u], vis[v]);
}
if(vis[u] == low[u])
{
scc_cnt++;
for(;;)
{
int x = S.top(); S.pop();
sccno[x] = scc_cnt;
if(x == u) break;
}
}
} void build(int mid)
{
for(int i = ; i <= mid; i++)
{
if(c[i] == )
{
G[a[i] << | ].push_back(b[i] << );
G[b[i] << | ].push_back(a[i] << );
}
else if(c[i] == )
{
G[a[i] << | ].push_back(b[i] << | );
G[b[i] << | ].push_back(a[i] << | );
G[a[i] << ].push_back(b[i] << );
G[b[i] << ].push_back(a[i] << );
}
else if(c[i] == )
{
G[a[i] << ].push_back(b[i] << | );
G[b[i] << ].push_back(a[i] << | );
}
}
} bool check()
{
for(int i = ; i < n * ; i += )
if(sccno[i] == sccno[i + ])
return false;
return true;
} int main()
{
int T;
rd(T);
while(T--)
{
init();
rd(n), rd(m);
for(int i = ; i < m; i++)
{
rd(a[i]), rd(b[i]), rd(c[i]);
}
int l = , r = m;
while(l + < r)
{
init();
int mid = (l + r) / ;
build(mid);
for(int i = ; i < n * ; i++)
if(!vis[i]) dfs(i);
if(check()) l = mid;
else r = mid;
}
pd(l + ); } return ;
}
Go Deeper
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3435 Accepted Submission(s): 1125
go(int dep, int n, int m)
begin
output the value of dep.
if dep < m and x[a[dep]] + x[b[dep]] != c[dep] then go(dep + 1, n, m)
end
In this code n is an integer. a, b, c and x are 4 arrays of integers. The index of array always starts from 0. Array a and b consist of non-negative integers smaller than n. Array x consists of only 0 and 1. Array c consists of only 0, 1 and 2. The lengths of array a, b and c are m while the length of array x is n. Given the elements of array a, b, and c, when we call the procedure go(0, n, m) what is the maximal possible value the procedure may output?
2 1
0 1 0
2 1
0 0 0
2 2
0 1 0
1 1 2
1
2
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