codeforces Yet Another DAG Problem

最新推荐文章于 2024-10-30 22:27:34 发布

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本文介绍了一道Codeforces竞赛题目的两种解法：一种是状态压缩动态规划，复杂度较高；另一种是通过构建最小割模型求解最优方案。文章提供了完整的代码实现，并对比了两者的优劣。

题目链接：https://codeforces.com/contest/1430/problem/G
这个题目可以采用状态压缩DP求解，最简单的是O(n*3^n)的暴力状态压缩DP，但是复杂度是不允许的，优化的也不会。这里参考了这篇博客，采用了最小割的建模方式，求解这个题目（https://www.luogu.com.cn/problem/solution/CF1430G）

这里提供一个O(n*3^n)的状态压缩DP

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)

template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}

int n , m;
const int N = 2e5 + 5;
int ve[1<<20];
ll a[1<<20], sum[1<<20];
vector <int> edge[N];
int dp[20][1<<20], f[20][1<<20], ans[20];
int main()
{
    read (n, m);
    for (int i = 1 ; i <= m ; i ++)
    {
        int u , v, w;
        read (u, v , w);
        u --;
        v --;
        edge[u].pb (v);
        a[u] += w;
        a[v] -= w;
    }
    for (int i = 0 ; i < (1 << n) ; i ++)
    {
        for (int j = 0 ; j < n ; j ++)
        {
            if ((i >> j) & 1)
                sum[i] += a[j];
        }
        for (int j = 0 ; j < n ; j ++)
        {
            if ((i >> j) & 1)
                for (auto v : edge[j])
                    ve[i] |= (1 << v);
        }
    }
    mem (dp, INF);
    dp[0][0] = 0;
    for (int i = 1 ; i <= n ; i ++)
    {
        for (int s = 0 ; s < (1<<n) ; s ++)
        {
            for (int s1 = s ; s1 ; s1 = (s1 - 1) & s)
            {
                int s2 = (s ^ s1);
                if ((ve[s1] & s2) == ve[s1])
                {
                    int cost = dp[i-1][s2] + sum[s1] * (i - 1);
                    if (cost < dp[i][s])
                    {
                        dp[i][s] = cost;
                        f[i][s] = s1;
                    }
                }
            }
        }
    }
    int s = (1 << n) - 1;
    int index = n;
    for (int i = 1 ; i <= n ; i ++)
    {
        if (dp[i][s] < dp[index][s])
            index = i;
    }
    for (int i = index ; i >= 1 ; i --)
    {
        int s1 = f[i][s];
        for (int j = 0 ; j < n ; j ++)
        {
            if ((s1 >> j) & 1)
                ans[j] = i - 1;
        }
        s = s ^ s1;
    }
    for (int i = 0 ; i < n ; i ++) write (ans[i]), SP;
}

这里是最小割：

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int ll
#define mem(a, b) memset(a,b,sizeof(a))
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)

template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}

const int N = 1e6 + 5;
int id[25][25];
int head[N] , cnt = 0 ,  tot = 0;
struct node {
    int t, c, next;
}edge[N];
void add (int f, int t, int c)
{
    edge[cnt].t = t;
    edge[cnt].c = c;
    edge[cnt].next = head[f];
    head[f] = cnt ++;
}
void addedge (int f, int t, int c)
{
    add (f, t, c);
    add (t, f, 0);
}
const int INF = 1ll<<60;
int n , m , ed , st, cur[N], level[N];
ll a[N];
int vis[N];
void dfs2 (int u)
{
    vis[u] = 1;
    for (int i = head[u] ; i != -1 ; i = edge[i].next)
    {
        if (edge[i].c && !vis[edge[i].t])
            dfs2 (edge[i].t);
    }
}
bool DinicBfs(int s, int e) //利用广搜给网络图分层
{
    mem(level,0);
    queue<int>q;
    q.push(s);
    level[s] = 1;
    while(!q.empty())
    {
        int root = q.front();
        q.pop();
        for(int i = head[root] ; i != -1 ; i = edge[i].next)
        {
            int v = edge[i].t, c = edge[i].c;
            if(!level[v] && c > 0)
            {
                level[v] = level[root] + 1;
                q.push(v);
            }
        }
    }
    return level[e] != 0; //如果当前网络能到达汇点说明存在增广路
}
int DinicDfs(int root , int flow) //利用DFS在已经分好层的网络中寻找增广路
{
    if(root == ed)
        return flow;
    int newflow , cost = 0;
    for(int& i = cur[root] ; i != -1 ; i = edge[i].next) //当前弧优化，&i代表i改动cur也会改动
    {
        int v = edge[i].t, c = edge[i].c;
        if(level[v] == level[root] + 1 && c > 0 && (newflow = DinicDfs(v,min(c,flow-cost))))//多路增广flow-cost
        {
            cost += newflow;
            edge[i].c -= newflow;
            edge[i^1].c += newflow;
            if(cost == flow) break;
        }
    }
    return cost; //返回当前分层网络中能够寻找到的最大流
}
int Dinic()
{
    int ans = 0;
    while (DinicBfs(st, ed))
    {
        for (int i = st ; i <= tot ; i ++) //如果顶点标号为0，则从0开始
            cur[i] = head[i];
        while (int add = DinicDfs(st, INF))
            ans += add;
    }
    return ans;
}
signed  main()
{
    mem (head, -1);
    cin >> n >> m;
    st = ++ tot , ed = ++ tot;
    for (int i = 1 ; i <= n ; i ++)
        for (int j = 0 ; j <= n ; j ++)
            id[i][j] = ++ tot;
    for (int i = 1 ;i  <= m ; i ++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        a[u] += w;
        a[v] -= w;
        for (int j = 0 ; j < n ; j ++)
            addedge (id[v][j], id[u][j+1], INF);
    }
    for (int i = 1 ; i <= n ; i ++)
    {
        addedge (st, id[i][0], INF);
        addedge (id[i][n], ed, INF);
        for (int j = 0 ; j < n ; j ++)
            addedge (id[i][j] , id[i][j+1], a[i] * j + 4e11);
    }
    Dinic();
    dfs2(st);
    for (int i = 1 ; i <= n ; i ++)
    {
        for (int j = n ; j >= 0 ; j --)
            if (vis[id[i][j]])
            {
                cout << j << ' ';
                break;
            }
    }
    cout << endl;
}