【网络流】 HDU 4309 Seikimatsu Occult Tonneru 状压枚举边

最新推荐文章于 2019-07-18 21:14:18 发布

原创最新推荐文章于 2019-07-18 21:14:18 发布 · 685 阅读

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本文介绍了一个涉及多种特殊路径约束的问题解决方法，通过使用二进制枚举和改进的SAP算法来寻找能够容纳最多人同时花费最少成本的路径。特别针对类型3路径进行了优化处理。

题意：有三种类型的路

类型 1. 从 U 到 V 能走INF个人能躲 d 个人

类型 2. 从 U 到 V 能走INF 个人不能躲人

类型 3. 从 U 到 V 能走INF 个人要用该路时需要花费一定费用

要求能躲最多人且花费最少

类型3 的路最多有 12 条

所以我们用二进制枚举类型3

（也有递归来枚举，dfs心太累）

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <time.h>;
#define cler(arr, val)    memset(arr, val, sizeof(arr))
#define FOR(i,a,b)  for(int i=a;i<=b;i++)
#define IN   freopen ("in.txt" , "r" , stdin);
#define OUT  freopen ("out.txt" , "w" , stdout);
typedef long long  LL;
const int MAXN = 200;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
//const int INF = 922222;
const int mod = 1000000007;
struct Edge
{
    int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof (head));
}
void addedge (int u,int v,int w,int rw = 0)
{
    edge[tol].to = v;
    edge[tol].cap = w;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = rw;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0] = 1;
    int front = 0, rear = 0;
    dep[end] = 0;
    Q[rear++] = end;
    while(front != rear)
    {
        int u = Q[front++];
        for(int i = head[u]; i !=  -1; i = edge[i].next)
        {
            int v = edge[i]. to;
            if(dep[v] != -1)continue;
            Q[rear++] = v;
            dep[v] = dep[u] + 1;
            gap[dep[v]]++;
        }
    }
}
int S[MAXN];
int sap(int start,int end, int N)
{
    BFS(start,end);
    memcpy(cur,head,sizeof(head));
    int top = 0;
    int u = start;
    int ans = 0;
    int i;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            int inser;
            for( i = 0; i < top; i++)
            {
                if(Min > edge[S[i]].cap - edge[S[i]].flow)
                {
                    Min = edge[S[i]].cap - edge[S[i]].flow;
                    inser = i;
                }
            }
            for( i = 0; i < top; i++)
            {
                edge[S[i]]. flow += Min;
                edge[S[i]^1].flow -= Min;
            }
            ans += Min;
            top = inser;
            u = edge[S[top]^1].to;
            continue;
        }
        bool flag =  false;
        int v;
        for( i = cur[u]; i != -1; i = edge[i]. next)
        {
            v = edge[i]. to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag =  true;
                cur[u] = i;
                break;
            }
        }
        if(flag)
        {
            S[top++] = cur[u];
            u = v;
            continue;
        }
        int Min = N;
        for( i = head[u]; i !=  -1; i = edge[i].next)
        {
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        }
        gap[dep[u]]--;
        if(!gap[dep[u]]) return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if(u != start)u = edge[S[--top]^1].to;
    }
    return ans;
}
int u[MAXM],v[MAXM],d[MAXM],p[MAXM],man[MAXN];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
#endif
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int star=0,end=n+1,mansum,num3=0,num1=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&man[i]);
            if(d!=0)
                mansum+=man[i];
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d%d",&u[i],&v[i],&d[i],&p[i]);
            if(p[i]<0&&d[i]!=0)
                num1++;
            else if(p[i]>0)
                num3++;
        }
        if(!num1)//优化了 下快了很多
        {
            puts("Poor Heaven Empire");
            continue;
        }
        int maxflow=0,cost=0;
        for(int k=0;k<(1<<num3);k++)//枚举类型3 
        {
            init();
            int x=end,cc=0,cos=0,ans;
            for(int i=1;i<=n;i++)
                addedge(star,i,man[i]);
            for(int i=0;i<m;i++)
            {
                if(p[i]<0)
                {
                    addedge(u[i],v[i],INF);
                    addedge(u[i],end,d[i]);
                }
                else if(p[i]==0)
                {
                    addedge(u[i],v[i],INF);
                }
                else if(p[i]>0)
                {
                    if(k&(1<<cc))
                    {
                        addedge(u[i],v[i],INF);
                        cos+=d[i];
                    }
                    else addedge(u[i],v[i],1);
                    cc++;
                }
            }
            ans=sap(star,end,n+2);
            if(ans>maxflow)
                maxflow=ans,cost=cos;
            else if(ans==maxflow&&cost>cos)
                cost=cos;
        }
        if(maxflow==0)
            puts("Poor Heaven Empire");
        else printf("%d %d\n",maxflow,cost);
    }
    return 0;
}

所以我们用二进制枚举类型3