网络流24题 软件补丁问题

Posted 2020-12-23 aaddvvaanntteezz

题目链接

题解

• 这题好像不是网络流?
• 注意到$n$最大只有20,所以可以考虑把$bug$的状态压缩成一个整数,对应位上为1代表存在这个$bug$,0表示不存在这个$bug$
• 然后就可以根据补丁建图跑最短路就行了

查看代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> p;
const int maxn = 3e6+5;
const int mod = 1e9+7;
ll qpow(ll a,ll b){ll res=1;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
struct graph
{
    int head[maxn],nxt[maxn<<1],to[maxn<<1],w[maxn<<1],sz;
    void init(){memset(head,-1,sizeof(head));}
    graph(){init();}
    void push(int a,int b,int c){nxt[sz]=head[a],to[sz]=b,w[sz]=c,head[a]=sz++;}
    int& operator[](const int a){return to[a];}
}g;
char s[22],t[22];
struct edge
{
    int need,noneed,creat,del,c;
}a[105];
int d[1100005];
bool vis[1100005];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("simple.in", "r", stdin);
    freopen("simple.out", "w", stdout);
#endif
    int n,m;
    scanf("%d%d",&n,&m);
    int mx = 1<<n;
    for(int i = 1,c;i <= m;++i){
        scanf("%d%s%s",&a[i].c,s,t);
        for(int j = 0;j < n;++j){
            if(s[j]==‘+‘)a[i].need|=1<<j;
            else if(s[j]==‘-‘)a[i].noneed|=1<<j;
            if(t[j]==‘+‘)a[i].creat|=1<<j;
            else if(t[j]==‘-‘)a[i].del|=1<<j;
        }
    }   
    memset(d,0x3f,sizeof(d));
    d[mx-1]=0;
    priority_queue<p>q;
    q.push(make_pair(0,mx-1));
    while(!q.empty()){
        p now = q.top();
        q.pop();
        int cost = -now.first;
        int u = now.second;
        if(cost>d[u])continue;
        vis[u]=1;
        for(int i = 1;i <= m;++i){
            if(((a[i].need&u)==a[i].need)&&((a[i].noneed&u)==0)){
                int tmp = u|a[i].del;
                tmp^=a[i].del;
                tmp|=a[i].creat;
                if(!vis[tmp]){
                    if(d[tmp]>cost+a[i].c){
                        d[tmp]=cost+a[i].c;
                        q.push(make_pair(-d[tmp],tmp));
                    }
                }
            }
        }
    }
    if(!vis[0]){
        printf("0
");
    }
    else printf("%d
",d[0]);
    return 0;   
}