[IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2)] -D. Delivery Bears （二分+最大流）
題面：
題意：
給定一個含有\(\mathit n\)個節點\(\mathit m\)個有向邊的圖和\(\mathit x\)個工作小熊。保證有一個\(1->n\)的路徑。
現在讓你選擇一個最大的運輸總值，滿足當這\(\mathit x\)個小熊均分任務後選擇一些路徑從\(1->n\)，使其可以滿足
每一個有向邊的容量不小於走過該邊的小熊的任務總量。
思路：
顯然可以在區間\([1e(-6),1e6]\)這個區間去二分出每一個小熊最大的運輸任務。
對於check當前二分的數值\(mid\)，我們需要用到網絡流中的最大流算法，
在原圖的基礎上將邊權\(c_i\)改爲\(c_i/mid\)向下取整的結果，跑最大流算法，得出\(1->n\)的最大流量\(res\)，
判斷\(res\)與\(\mathit x\)的關係即可知道轉移的方向。
我才用ISAP算法求最大流：該算法的時間複雜度上限：\(O(E*V^2)\)，理想情況：\(O(sqrt(E)*V^2)\)
於是本題的ac代碼時間複雜度爲\(O(m*n^2*log_2x)\)
代碼：
#include<bits/stdc++.h>
using namespace std;
#define N 1000
#define INF 1e15
typedef long long ll;
struct Edge {
int from, to;
ll cap, flow;
};
struct ISAP {
int n, m, s, t;
std::vector<Edge> edges;
std::vector<int> G[N];
bool vis[N];
int d[N], cur[N];
int p[N], num[N];
void addedge(int from, int to, ll cap)
{
edges.push_back((Edge) {from, to, cap, 0});
edges.push_back((Edge) {to, from, 0, cap});
int m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
void init()
{
memset(d, 0, sizeof(d));
edges.clear();
for (int i = 0; i <= n; ++i) {
G[i].clear();
}
}
ll Augument()
{
ll x = t, a = INF;
while (x != s) {
Edge &e = edges[p[x]];
a = min(a, e.cap - e.flow);
x = edges[p[x]].from;
}
x = t;
while (x != s) {
edges[p[x]].flow += a;
edges[p[x] ^ 1].flow -= a;
x = edges[p[x]].from;
}
return a;
}
void bfs()
{
memset(vis, 0, sizeof(vis));
queue<int> q;
q.push(t);
d[t] = 0;
vis[t] = 1;
while (!q.empty()) {
int x = q.front();
q.pop();
int len = G[x].size();
for (int i = 0; i < len; ++i) {
Edge &e = edges[G[x][i]];
if (!vis[e.from] && e.cap > e.flow) {
vis[e.from] = 1;
d[e.from] = d[x] + 1;
q.push(e.from);
}
}
}
}
ll Maxflow(int s, int t)
{
this->s = s;
this->t = t;
ll flow = 0;
bfs();
memset(num, 0, sizeof(num));
for (int i = 0; i < n; ++i) {
num[d[i]]++;
}
int x = s;
memset(cur, 0, sizeof(cur));
while (d[s] < n) {
if (x == t) {
flow += Augument();
x = s;
}
int ok = 0;
for (int i = cur[x]; i < G[x].size(); ++i) {
Edge &e = edges[G[x][i]];
if (e.cap > e.flow && d[x] == d[e.to] + 1) {
ok = 1;
p[e.to] = G[x][i];
cur[x] = i;
x = e.to;
break;
}
}
if (!ok) {
int m = n - 1;
for (int i = 0; i < G[x].size(); ++i) {
Edge &e = edges[G[x][i]];
if (e.cap > e.flow) {
m = min(m, d[e.to]);
}
}
if (--num[d[x]] == 0) {
break;
}
num[d[x] = m + 1]++;
cur[x] = 0;
if (x != s) {
x = edges[p[x]].from;
}
}
}
return flow;
}
} gao;
int n;
int m;
int x;
int a[N], b[N], c[N];
bool check(double mid)
{
int S = 1;
int T = n;
gao.n = T + 1;
for (int i = 1; i <= m; ++i) {
ll num = floor(1.0 * c[i] / mid);
gao.addedge(a[i], b[i], num);
}
int res = gao.Maxflow(S, T);
gao.init();
return res >= x;
}
int main()
{
scanf("%d %d %d", &n, &m, &x);
for (int i = 1; i <= m; ++i) {
scanf("%d %d %d", &a[i], &b[i], &c[i]);
}
double l = 1e-6;
double r = 1000000;
double ans, mid;
for(int rp=1;rp<=100;++rp)
{
mid = (l + r) * 0.5;
if (check(mid)) {
l = mid;
ans = mid;
} else {
r = mid;
}
}
ans*=x;
printf("%.6f\n", ans );
return 0;
}