C++ Program for Fedor’s Roadmap Navigation Problem
- Fedor is a research scientist, who has recently found a roadmap of Ancient Berland.
- Ancient Berland consisted of N cities that were connected by M bidirectional roads. The road builders weren’t knowledgable. Hence, the start city and the end city for each road were always chosen randomly and independently. As a result, there were more than one road between some pairs of cities. Nevertheless, by luck, the country remained connected (i.e. you were able to get from one city to another via these M roads). And for any road, the start and the end city were not the same.
- Moreover, each road had it’s own value of importance. This value was assigned by the Road Minister of Ancient Berland. The Road Minister also was not knowledgable, so these numbers were assigned to the roads randomly and independently from the other roads.
- When there was a war with the neighboring countries (usually it was with Ancient Herland), it was important to estimate separation number for some pairs of cities.
- The separation number for a pair of cities – let’s call these cities A and B – is explained below:
Consider a set of roads that were built. The subset of this set is good, if after removing all roads from this set, there’s no longer a way from A to B. The minimal possible sum of roads’ value of importance of any good subset is a separation number for the pair of cities (A, B).
For a research, Fedor would like to know the product of separation values over all unordered pairs of cities. Please, find this number. It can be huge, so we ask you to output its product modulo 109+7.
Explanation
Input Format
The first line of input consist of two integers N and M, separated by a single space.
Then, M lines follow. Each of these lines consist of three integers Xi, Yi, Zi separated by a single space.
It means that there was a road between the city Xi and the city Yi with a value of importance equal to Zi.
Constraints
3 ≤ N ≤ 500
3 ≤ M ≤ 104
1 ≤ value of importance ≤ 105
The cities are indexed from 1 to N.
Output Format
An integer that represents the value, Fedor needs, modulo 109+7.
Sample Input
3 3
1 2 3
2 3 1
1 3 2
Sample Output
36
Explanation
There are three unordered pairs of cities: (1, 2), (1, 3) and (2, 3). Let’s look at the separation numbers:
For (1, 2) we have to remove the first and the second roads. The sum of the importance values is 4.
For (1, 3) we have to remove the second and the third roads. The sum of the importance values is 3.
For (2, 3) we have to remove the second and the third roads. The sum of the importance values is 3. So, we get 4 * 3 * 3 = 36.
Scoring
- In the 25% of the test data N = 50 and M = 300.
- In another 25% of the test data N = 200 and M = 10000
- In the rest of the test data N = 500 and M = 10000
Time Limit:3.0 sec(s) for each input file.
Memory Limit:256 MB
Source Limit:1024 KB
SOURCE CODE ::
#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <vector>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <cassert>
#include <numeric>
#include <iomanip>
#include <sstream>
#include <fstream>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<PII> VPII;
#define MM(a,x) memset(a,x,sizeof(a));
#define ALL(x) (x).begin(), (x).end()
#define P(x) cerr<<"["#x<<" = "<<(x)<<"]\n"
#define PP(x,i) cerr<<"["#x<<i<<" = "<<x[i]<<"]\n"
#define P2(x,y) cerr<<"["#x" = "<<(x)<<", "#y" = "<<(y)<<"]\n"
#define TM(a,b) cerr<<"["#a" -> "#b": "<<1e3*(b-a)/CLOCKS_PER_SEC<<"ms]\n";
#define UN(v) sort(ALL(v)), v.resize(unique(ALL(v))-v.begin())
#define mp make_pair
#define pb push_back
#define x first
#define y second
struct _ {_() {ios_base::sync_with_stdio(0);}} _;
template<class A, class B> ostream& operator<<(ostream &o, pair<A, B> t) {o << "(" << t.x << ", " << t.y << ")"; return o;}
template<class T> void PV(T a, T b) {while(a != b)cout << *a++, cout << (a != b ? " " : "\n");}
template<class T> inline bool chmin(T &a, T b) {return a > b ? a = b, 1 : 0;}
template<class T> inline bool chmax(T &a, T b) {return a < b ? a = b, 1 : 0;}
template<class T> string tostring(T x, int len = 0) {stringstream ss; ss << x; string r = ss.str(); if(r.length() < len) r = string(len - r.length(), '0') + r; return r;}
template<class T> void convert(string x, T& r) {stringstream ss(x); ss >> r;}
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3fLL;
const int mod = int(1e9) + 7;
const int N = 501;
#define FF int
const int maxE = 40010;
const int maxN = 10010;
const int INF = 0x3f3f3f3f;
struct Dinic {
int ne;
int hd[maxN], work[maxN], q[maxN], Level[maxN], from[maxE], to[maxE], next[maxE];
FF cap[maxE], flow[maxE];
Dinic() {init();}
void init() {
ne = 0;
memset(hd, -1, sizeof(hd));
}
void add(int x, int y, FF c) {
from[ne] = x, to[ne] = y, cap[ne] = c, flow[ne] = 0, next[ne] = hd[x], hd[x] = ne++;
from[ne] = y, to[ne] = x, cap[ne] = 0, flow[ne] = 0, next[ne] = hd[y], hd[y] = ne++;
}
void addU(int x, int y, FF c) {
from[ne] = x, to[ne] = y, cap[ne] = c, flow[ne] = 0, next[ne] = hd[x], hd[x] = ne++;
from[ne] = y, to[ne] = x, cap[ne] = c, flow[ne] = 0, next[ne] = hd[y], hd[y] = ne++;
}
bool dinic_bfs(int S, int T) {
int head = 0, tail = 0;
memset(Level, 0, sizeof(Level));
Level[S] = 1;
q[tail++] = S;
while(head < tail) {
int u = q[head++];
for(int i = hd[u]; i != -1; i = next[i]) {
int v = to[i];
if(flow[i] < cap[i] && !Level[v]) {
Level[v] = Level[u] + 1;
q[tail++] = v;
if(v == T) return 1;
}
}
}
return 0;
}
FF dinic_dfs(int u, int T, FF pMin) {
if(u == T || !pMin) return pMin;
FF ret = 0;
for(int& i = work[u]; i != -1; i = next[i]) {
int v = to[i];
FF f;
if(Level[v] == Level[u] + 1 && (f = dinic_dfs(v, T, min(pMin, cap[i] - flow[i])))) {
flow[i] += f;
flow[i ^ 1] -= f;
ret += f;
pMin -= f;
if(pMin == 0) break;
}
}
return ret;
}
FF dinic(int S, int T) {
FF ret = 0;
while(dinic_bfs(S, T)) {
memcpy(work, hd, sizeof(hd));
ret += dinic_dfs(S, T, INF);
}
return ret;
}
} dn;
int n, m;
vector<pair<PII, int>> E;
int visited[N];
void dfs(int cur, vector<vector<int>>& G) {
if(visited[cur]) return;
visited[cur] = 1;
for(auto i : G[cur]) dfs(i, G);
}
int parent[N];
int cut[N][N];
int main() {
cin >> n >> m;
assert(3 <= n && n <= 500);
assert(3 <= m && m <= 1e4);
for(int i = 1; i <= m; i++) {
int u, v, w;
cin >> u >> v >> w;
assert(1 <= u && u <= n);
assert(1 <= v && v <= n);
assert(u != v);
assert(1 <= w && w <= 1e5);
u--, v--;
E.pb(mp(mp(u, v), w));
}
MM(cut, 0x3f);
for(int i = 1; i < n; i++) {
dn.init();
for(auto e : E) dn.addU(e.x.x, e.x.y, e.y);
int S = i, T = parent[i];
int x = dn.dinic(S, T);
MM(visited, 0);
vector<vector<int>> G(n + 1);
for(int j = 0; j < dn.ne; j++) if(dn.cap[j] > dn.flow[j]) G[dn.from[j]].pb(dn.to[j]);
dfs(S, G);
for(int j = i + 1; j < n; j++) if(visited[j] && parent[j] == parent[i]) parent[j] = i;
cut[i][parent[i]] = cut[parent[i]][i] = x;
for(int j = 0; j < i; j++) {
cut[i][j] = cut[j][i] = min(x, cut[parent[i]][j]);
}
}
LL res = 1;
for(int i = 0; i < n; i++)
for(int j = i + 1; j < n; j++)
res = res * cut[i][j] % mod;
cout << res << endl;
return 0;
}
OUTPUT : :
3 3 1 2 3 2 3 1 1 3 2 36