Task: | Download Speed |
Sender: | minghao |
Submission time: | 2024-10-24 19:35:08 +0300 |
Language: | C++ (C++20) |
Status: | READY |
Result: | ACCEPTED |
test | verdict | time | |
---|---|---|---|
#1 | ACCEPTED | 0.00 s | details |
#2 | ACCEPTED | 0.01 s | details |
#3 | ACCEPTED | 0.00 s | details |
#4 | ACCEPTED | 0.00 s | details |
#5 | ACCEPTED | 0.01 s | details |
#6 | ACCEPTED | 0.01 s | details |
#7 | ACCEPTED | 0.01 s | details |
#8 | ACCEPTED | 0.00 s | details |
#9 | ACCEPTED | 0.00 s | details |
#10 | ACCEPTED | 0.00 s | details |
#11 | ACCEPTED | 0.01 s | details |
#12 | ACCEPTED | 0.00 s | details |
Compiler report
input/code.cpp: In member function 'LL EK::Maxflow(LL, LL)': input/code.cpp:49:26: warning: comparison of integer expressions of different signedness: 'LL' {aka 'long long int'} and 'std::vector<long long int>::size_type' {aka 'long unsigned int'} [-Wsign-compare] 49 | for (LL i = 0; i < G[x].size(); i++) { | ~~^~~~~~~~~~~~~
Code
#include<bits/stdc++.h> using namespace std; typedef long long LL; const LL N=1005, INF=0x7ffffffffffffff; struct Edge { LL from, to, cap, flow; Edge(LL u, LL v, LL c, LL f) : from(u), to(v), cap(c), flow(f) {} }; vector<vector<LL> > ans; struct EK { LL n, m; vector<Edge> edges; vector<LL> G[N]; LL a[N], p[N]; // a:点 x -> BFS 过程中最近接近点 x 的边给它的最大流 // p:点 x -> BFS 过程中最近接近点 x 的边 void init(LL n_) { n = n_; for (LL i = 0; i < n; i++) G[i].clear(); edges.clear(); } void AddEdge(LL from, LL to, LL cap) { edges.push_back(Edge(from, to, cap, 0)); edges.push_back(Edge(to, from, 0, 0)); m = edges.size(); G[from].push_back(m - 2); G[to].push_back(m - 1); } LL Maxflow(LL s, LL t) { LL flow = 0; while(true) { memset(a, 0, sizeof(a)); queue<LL> Q; Q.push(s); a[s] = INF; while (!Q.empty()) { LL x = Q.front(); Q.pop(); for (LL i = 0; i < G[x].size(); i++) { // 遍历以 x 作为起点的边 Edge& e = edges[G[x][i]]; if (!a[e.to] && e.cap > e.flow) { // G[x][i] 是最近接近点 e.to 的边 p[e.to] = G[x][i]; // 最近接近点 e.to 的边赋给它的流 a[e.to] = min(a[x], e.cap - e.flow); Q.push(e.to); } } if (a[t]) break; } // 如果汇点没有接受到流,说明源点和汇点不在同一个连通分量上 if (!a[t]) break; for (LL u = t; u != s; u = edges[p[u]].from) { // 通过 u 追寻 BFS 过程中 s -> t 的路径 edges[p[u]].flow += a[t]; edges[p[u] ^ 1].flow -= a[t]; } flow += a[t]; } return flow; } }graph; int main() { LL n, m; cin >> n >> m; graph.init(n); for(LL i=1; i<=m; i++) { LL a, b, c; cin >> a >> b >> c; graph.AddEdge(a, b, c); } cout << graph.Maxflow(1, n) << endl; return 0; }
Test details
Test 1
Verdict: ACCEPTED
input |
---|
4 3 1 2 5 2 3 3 3 4 6 |
correct output |
---|
3 |
user output |
---|
3 |
Test 2
Verdict: ACCEPTED
input |
---|
4 5 1 2 1 1 3 1 2 3 1 2 4 1 ... |
correct output |
---|
2 |
user output |
---|
2 |
Test 3
Verdict: ACCEPTED
input |
---|
4 5 1 2 1000000000 1 3 1000000000 2 3 1 2 4 1000000000 ... |
correct output |
---|
2000000000 |
user output |
---|
2000000000 |
Test 4
Verdict: ACCEPTED
input |
---|
2 1 2 1 100 |
correct output |
---|
0 |
user output |
---|
0 |
Test 5
Verdict: ACCEPTED
input |
---|
2 1000 1 2 1000000000 1 2 1000000000 1 2 1000000000 1 2 1000000000 ... |
correct output |
---|
1000000000000 |
user output |
---|
1000000000000 |
Test 6
Verdict: ACCEPTED
input |
---|
500 998 1 2 54 1 3 59 1 4 83 2 5 79 ... |
correct output |
---|
60 |
user output |
---|
60 |
Test 7
Verdict: ACCEPTED
input |
---|
500 998 1 2 530873053 1 3 156306296 1 4 478040476 3 5 303609600 ... |
correct output |
---|
1093765123 |
user output |
---|
1093765123 |
Test 8
Verdict: ACCEPTED
input |
---|
2 1 1 2 1 |
correct output |
---|
1 |
user output |
---|
1 |
Test 9
Verdict: ACCEPTED
input |
---|
4 5 1 2 3 2 4 2 1 3 4 3 4 5 ... |
correct output |
---|
6 |
user output |
---|
6 |
Test 10
Verdict: ACCEPTED
input |
---|
4 5 1 2 1 1 3 2 3 2 1 2 4 2 ... |
correct output |
---|
3 |
user output |
---|
3 |
Test 11
Verdict: ACCEPTED
input |
---|
10 999 1 2 1000000000 1 2 1000000000 1 2 1000000000 1 2 1000000000 ... |
correct output |
---|
111000000000 |
user output |
---|
111000000000 |
Test 12
Verdict: ACCEPTED
input |
---|
7 9 1 2 1 1 3 1 1 4 1 2 5 1 ... |
correct output |
---|
2 |
user output |
---|
2 |