最小路径覆盖算法解析-优快云博客

本文介绍了一种基于最小路径覆盖的问题解决方法，并提供了一个具体的实现案例。通过使用dinic算法来寻找最大流，进而确定最小路径覆盖。文章还包含完整的C++代码实现，展示了如何通过构造图模型来解决这一特定问题。

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https://loj.ac/problem/6003

变化了下的最小路径覆盖，要注意边数巨多。。。

#include <iostream>
#include <stdio.h>
#include <queue>
#include <cstring>
using namespace std;
int biao[100];
const int maxn = 10009;
int head[maxn];
int cur[maxn];
int tot = 0;
struct edge{
    int v,nex,w;
}e[maxn*20];
void addedge(int u,int v,int w){
    e[tot] = (edge){v,head[u],w};
    head[u] = tot++;
    e[tot] = (edge){u,head[v],w};
    head[v] = tot++;
}
int deep[maxn];
int nowball;
bool bfs(int S,int T){
    for(int i=0;i<=2*nowball+1;i++){
        deep[i] = 0;
    }
    queue<int> q;
    deep[S] = 1;
    q.push(S);
    while(!q.empty()){
        int now = q.front();
        q.pop();
        for(int i=head[now];i!=-1;i=e[i].nex){
            int v = e[i].v;
            int w = e[i].w;
            if(deep[v]!=0 || w<=0) continue;
            deep[v] = deep[now]+1;
            if(v==T) return deep[T];
            q.push(v);
        }
    }
    return deep[T];
}
int dfs(int now,int T,int maxflow){
    if(now==T) return maxflow;
    int all  =0;
    for(int &i=cur[now];i!=-1 &&all <maxflow;i=e[i].nex){
        int v = e[i].v;
        int w = e[i].w;
        if(deep[v]!=deep[now]+1 ||w<=0) continue;
        int tt = dfs(v,T,min(maxflow-all,w));
        e[i].w-=tt;
        e[i^1].w+=tt;
        all+=tt;
    }
    return all;
}

int dinic(int S,int T){
    int ret = 0;
    for(int i=0;i<=2*nowball+1;i++){
        cur[i] = head[i];
    }
    while(bfs(S,T)){
        ret+=dfs(S,T,0x3f3f3f3f);
        for(int i=0;i<=2*nowball;i++){
            cur[i] = head[i];
        }
    }
    return ret;
}
int vis[maxn*2];
void  print(int now){
    if(now<=1 || now%2==1) return;
    vis[now] = 1;
    printf("%d ",now/2);
    for(int i=head[now];i!=-1;i=e[i].nex){
        int v = e[i].v;
        int w = e[i].w;
        if(w<=0){
            print(v^1);
        }
    }
}
int main()
{
    for(int i=0;i<100;i++){
        biao[i] = i*i;
    }
    memset(head,-1,sizeof(head));
    int n;
    scanf("%d",&n);
    int kkk = 0;
    nowball = 0;
    while(true){
        nowball++;
        for(int i=0;i<100;i++){
            int from = biao[i]-nowball;
            if(from<1 || from>=nowball) continue;
            addedge(from*2,nowball*2+1,1);
        }
        addedge(0,nowball*2,1);
        addedge(nowball*2+1,1,1);
        kkk+=dinic(0,1);
        if(nowball-kkk<=n) continue;
        else{
            break;
        }
    }
    nowball--;
    printf("%d\n",nowball);
    for(int i=1;i<=nowball;i++){
        if(vis[2*i]==0) {
            print(2*i);
            printf("\n");
        }

    }
    return 0;
}

转载于:https://www.cnblogs.com/tjucxz/p/8562694.html