洛谷 3305 「SDOI2013」费用流

过分显然的是，Bob 一定是把 P 全都分配在流量最大的那条边上……

也就是在保证最大流的前提下，要流量最大的边的流量最小……
二分一下答案（需要实数），与原来的容量取个 min 跑最大流。

代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100;
const int M = 1e3;
const double eps = 1e-5;
int n,m,p,s,t;
struct Edge
{
    int u,v,w;
} e[M + 5];
int to[(M << 1) + 5],pre[(M << 1) + 5],first[N + 5],cur[N + 5],edge_tot;
double val[(M << 1) + 5];
inline void add(int u,int v,double w)
{
    int &tot = edge_tot;
    to[tot] = v,val[tot] = w,pre[tot] = first[u],first[u] = tot++;
}
int head,tail,q[N + 5],dep[N + 5],vis[N + 5];
int bfs()
{
    head = tail = 0;
    memset(vis,0,sizeof vis);
    memset(dep,0x3f,sizeof dep);
    dep[s] = 0,vis[s] = 1,q[++tail] = s;
    for(register int p;head < tail;)
    {
        vis[p = q[++head]] = 0;
        for(register int i = first[p];~i;i = pre[i])
            if(val[i] && dep[to[i]] > dep[p] + 1)
            {
                dep[to[i]] = dep[p] + 1;
                if(!vis[to[i]])
                    vis[to[i]] = 1,q[++tail] = to[i];
            }
    }
    return dep[t] ^ 0x3f3f3f3f;
}
double dfs(int p,double flow)
{
    if(p == t)
        return flow;    
    double ret = 0,used = 0;
    for(register int i = cur[p];~i;cur[p] = i = pre[i])
        if(val[i] > 0 && dep[to[i]] == dep[p] + 1 && (ret = dfs(to[i],min(flow - used,val[i]))) > 0)
        {
            used += ret,val[i] -= ret,val[i ^ 1] += ret;
            if(used >= flow)
                break;
        }
    return used;
}
double dinic()
{
    double ret = 0;
    while(bfs())
    {
        for(register int i = 1;i <= n;++i)
            cur[i] = first[i];
        ret += dfs(s,0x3f3f3f3f);
    }
    return ret;
}
int maxflow;
double l,r,mid,ans;
int check()
{
    memset(first,-1,sizeof first),edge_tot = 0;
    for(register int i = 1;i <= m;++i)
        add(e[i].u,e[i].v,min((double)e[i].w,mid)),add(e[i].v,e[i].u,0);
    double flow = dinic();
    return fabs(flow - maxflow) < eps;
}
int main()
{
    memset(first,-1,sizeof first);
    scanf("%d%d%d",&n,&m,&p),s = 1,t = n;
    for(register int i = 1;i <= m;++i)
        scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w),add(e[i].u,e[i].v,e[i].w),add(e[i].v,e[i].u,0);
    r = maxflow = dinic();
    for(register int i = 1;i <= 100;++i)
        mid = (l + r) / 2,check() ? (ans = r = mid) : (l = mid);
    printf("%d\n%.4f\n",maxflow,ans * p);
}