zoj-3792-Romantic Value-最小割+数值转化

标签：class   blog   code   ext   2014   string   

如果不需要求边的个数的话，就是一个裸的最小割问题。

求边的个数就用边的权值记录一下。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include<queue>
using namespace std;
#define INF 99999999
#define LL long long
const LL maxn =55;
const LL maxm =4400;
const LL oo = (LL)1<<37;
struct Arclist
{
    LL cnt, head[maxn], dis[maxn];
    LL cur[maxn], pre[maxn], gap[maxn], aug[maxn];
    struct node
    {
        LL u, v, w, next;
    }edge[maxm];
    void init()
    {
        cnt = 0;
        memset(head,-1,sizeof(head));
    }
    void add(LL u, LL v, LL w)
    {
       // cout<<u<<" "<<v<<" "<<w<<endl;
        edge[cnt].u = u;
        edge[cnt].v = v;
        edge[cnt].w = w;
        edge[cnt].next = head[u];
        head[u] = cnt++;
        edge[cnt].u = v;
        edge[cnt].v = u;
        edge[cnt].w = 0;
        edge[cnt].next = head[v];
        head[v] = cnt++;
    }
    LL sap(LL s, LL e, LL n)
    {
        LL max_flow = 0, u = s;
        LL mindis;
        for(LL i = 0; i <= n; i++)
        {
            cur[i] = head[i];
            dis[i] = 0;
            gap[i] = 0;
        }
        aug[s] = oo;
        pre[s] = -1;
        gap[0] = n;
        while(dis[s]<n)
        {
            bool flag = false;
            if(u==e)
            {
                max_flow += aug[e];
                for(LL v = pre[e]; v != -1; v = pre[v])
                {
                    LL id = cur[v];
                    edge[id].w -= aug[e];
                    edge[id^1].w += aug[e];
                    aug[v] -= aug[e];
                    if(edge[id].w==0) u = v;
                }
            }
            for(LL id = cur[u]; id != -1; id = edge[id].next)
            {
                LL v = edge[id].v;
                if(edge[id].w>0 && dis[u]==dis[v]+1)
                {
                    flag = true;
                    pre[v] = u;
                    cur[u] = id;
                    aug[v] = std::min(aug[u], edge[id].w);
                    u = v;
                    break;
                }
            }
            if(flag==false)
            {
                if(--gap[dis[u]]==0) break;
                mindis = n;
                cur[u] = head[u];
                for(LL id = head[u]; id != -1; id = edge[id].next)
                {
                    LL v = edge[id].v;
                    if(edge[id].w>0 && dis[v]<mindis)
                    {
                        mindis = dis[v];
                        cur[u] = id;
                    }
                }
                dis[u] = mindis+1;
                ++gap[dis[u]];
                if(u!=s) u = pre[u];
            }
        }
        return max_flow;
    }
}G;
int main()
{
    LL m,n,u,v,w,T,st,ed;
    scanf("%lld",&T);
    while(T--)
    {
        scanf("%lld%lld%lld%lld",&n,&m,&st,&ed);
        G.init();
        LL all=0;
        while(m--)
        {
            scanf("%lld%lld%lld",&u,&v,&w);
            G.add(v,u,w*1000+1);
            G.add(u,v,w*1000+1);
            all+=w;
        }
        LL w=G.sap(st,ed,n);
       // cout<<w<<endl;
        if(w==0)cout<<"Inf"<<endl;
        else
        {
            double x,y;
            if(w%1000==0)
            {
                y=1000;
                w=w-1000;
            }
            else y=w%1000;
            x=all-w/1000;
            printf("%.2f\n",1.0*x/y);
        }
    }
    return 0;
}

zoj-3792-Romantic Value-最小割+数值转化,布布扣,bubuko.com

zoj-3792-Romantic Value-最小割+数值转化

标签：class   blog   code   ext   2014   string   

原文地址：http://blog.csdn.net/rowanhaoa/article/details/32940079