Codeforces 653D Delivery Bears 【二分+最大流】

题目链接：Codeforces 653D Delivery Bears

题意：有n个点m条有向边带权图，权值代表该路径允许通过的最大物品重量。现在你要用x个小熊同时从点1到点n运物品，要求每个小熊运的物品重量相等。问所有小熊可以运送的最大重量和。

思路：不小心精度卡死我了。建图没什么好说的，新图的边权就是用路径的权值除以当前二分的值再向下取整。我们跑一发最大流，看是否满流。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#define PI acos(-1.0)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 500 + 10;
const int pN = 1e6;// <= 10^7
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
pii a[MAXN];
struct Edge {
int from, to, cap, flow, next;
};
Edge edge[MAXN*3];
int head[MAXN], cur[MAXN], edgenum;
void init() {CLR(head, -1); edgenum = 0;}
void addEdge(int u, int v, int w) {
Edge E1 = {u, v, w, 0, head[u]};
edge[edgenum] = E1;
head[u] = edgenum++;
Edge E2 = {v, u, 0, 0, head[v]};
edge[edgenum] = E2;
head[v] = edgenum++;
}
int dist[MAXN]; bool vis[MAXN];
bool BFS(int s, int t) {
queue<int> Q;
CLR(dist, -1); CLR(vis, false);
Q.push(s); vis[s] = true; dist[s] = 0;
while(!Q.empty())
{
int u = Q.front(); Q.pop();
for(int i = head[u]; i != -1; i = edge[i].next) {
Edge E = edge[i];
if(!vis[E.to] && E.cap > E.flow) {
vis[E.to] = true;
dist[E.to] = dist[u] + 1;
if(E.to == t) return true;
Q.push(E.to);
}
}
}
return false;
}
int DFS(int x, int a, int t)
{
if(x == t || a == 0) return a;
int flow = 0, f;
for(int &i = cur[x]; i != -1; i = edge[i].next) {
Edge &E = edge[i];
if(dist[E.to] == dist[x] + 1 && (f = DFS(E.to, min(E.cap - E.flow, a), t)) > 0) {
E.flow += f;
edge[i^1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int Maxflow(int s, int t) {
int flow = 0;
while(BFS(s, t)) {
memcpy(cur, head, sizeof(head));
flow += DFS(s, INF, t);
}
return flow;
}
int n, m, x;
int u[MAXN], v[MAXN], w[MAXN];
bool judge(double mid) {
init();
for(int i = 0; i < m; i++) {
addEdge(u[i], v[i], (int)min(x * 1.0, w[i] * 1.0 / mid));
}
//cout << Maxflow(1, n) << endl;
return Maxflow(1, n) >= x;
}
int main()
{
scanf("%d%d%d", &n, &m, &x);
for(int i = 0; i < m; i++) {
scanf("%d%d%d", &u[i], &v[i], &w[i]);
}
double l = 0, r = 1e10, ans = 0;
for(int i = 0; i <= 100; i++) {
double mid = (l + r) / 2;
if(judge(mid)) {
ans = mid;
l = mid;
}
else {
r = mid;
}
}
//cout << ans << endl;
printf("%.10lf\n", ans * x);
return 0;
}