一个神秘的oj2093 花园的守护之神（最小割）

给定一张无向图，你每次可以将一条路的权值增加1，询问最少增加多少次才会使得\(s->t\)的最短路改变

QwQ一看到这个题，我就用种最小割的感觉

我们可以把最短路上的点取出来，然后做最小割呀！！

首先

我们将最短路求一下\(dis[i]\)表示\(s\)到\(i\)的最短距离，\(disn[i]\)表示\(t\)到\(i\)的最短路。

如果一条边\(u->v\)

满足\(dis[u]+val[i]+disn[v]==dis[t]\)

那么他就是最短路上的边了。

这里注意要将双向边看成两个单向边来做，不然会出bug

上代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#define pa pair<long long,long long>
#define ll long long
using namespace std;

inline int read()
{
  int x=0,f=1;char ch=getchar();
  while (!isdigit(ch)) {if (ch==‘-‘) f=-1;ch=getchar();}
  while (isdigit(ch)) {x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();}
  return x*f;
}

inline ll read1()
{
  ll x=0,f=1;char ch=getchar();
  while (!isdigit(ch)) {if (ch==‘-‘) f=-1;ch=getchar();}
  while (isdigit(ch)) {x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();}
  return x*f;
}

const int maxm = 2e6+1e2;
const int maxn = 1010;
const int inf = 2e9;

int point[maxn],nxt[maxm],to[maxm];
ll val[maxm];
int cnt,n,m;
int x[maxm],y[maxm];
ll w[maxm];
ll dis[maxn];
int vis[maxn];
ll disn[maxn];
int h[maxn];
int s,t;
queue<int> q;
priority_queue<pa,vector<pa>,greater<pa> > que;

void addedge(int x,int y,ll w)
{
    nxt[++cnt]=point[x];
    to[cnt]=y;
    val[cnt]=w;
    point[x]=cnt;
}

void insert(int x,int y,ll w)
{
    addedge(x,y,w);
    addedge(y,x,0);
}

void init()
{
  cnt=1;
  memset(point,0,sizeof(point));
}

bool bfs(int s)
{
    memset(h,-1,sizeof(h));
    h[s]=0;
    q.push(s);
    while (!q.empty())
    {
        int x = q.front();
        q.pop();
        for (int i=point[x];i;i=nxt[i])
        {
            int p = to[i];
            if (val[i]>0 && h[p]==-1)
            {
                h[p]=h[x]+1;
                q.push(p);
            }
        }
    }
    if (h[t]==-1) return 0;
    else return 1;
}

int dfs(int x,int low)
{
    if (x==t || low==0) return low;
    int totflow=0;
    for (int i=point[x];i;i=nxt[i])
    {
        int p = to[i];
        if (val[i]>0 && h[p]==h[x]+1)
        {
            int tmp = dfs(p,min(low,(int)val[i]));
            val[i]-=tmp;
            val[i^1]+=tmp;
            low-=tmp;
            totflow+=tmp;
            if (low==0) return totflow;
        }
     }
     if (low>0) h[x]=-1;
     return totflow;
}

int dinic()
{
    int ans=0;
    while (bfs(s))
    {
        ans=ans+dfs(s,inf);
    }
    return ans;
}

void dijkstra(int s)
{
    memset(vis,0,sizeof(vis));
    memset(dis,127/3,sizeof(dis));
    dis[s]=0;
    //vis[s]=1;
    que.push(make_pair(0,s));
    while (!que.empty())
    {
        int x = que.top().second;
        que.pop();
        if (vis[x]) continue;
        vis[x]=1;
        for (register int i=point[x];i;i=nxt[i])
        {
            int p = to[i];
            if (dis[p]>dis[x]+val[i])
            {
                dis[p]=dis[x]+val[i];
                que.push(make_pair(dis[p],p));
            }
        }
    }
}

void dijkstran(int s)
{
    memset(vis,0,sizeof(vis));
    memset(disn,127/3,sizeof(disn));
    disn[s]=0;
    //vis[s]=1;
    que.push(make_pair(0,s));
    while (!que.empty())
    {
        int x = que.top().second;
        que.pop();
        if (vis[x]) continue;
        vis[x]=1;
        for (register int i=point[x];i;i=nxt[i])
        {
            int p = to[i];
            if (disn[p]>disn[x]+val[i]){
                disn[p]=disn[x]+val[i];
                que.push(make_pair(disn[p],p));
            }
        }
    }
}

int main()
{
  freopen("greendam2002.in","r",stdin);
  freopen("greendam2002.out","w",stdout);
  scanf("%d%d%d%d",&n,&m,&s,&t);
  for (register int i=1;i<=m;i++) x[i]=read(),y[i]=read(),w[i]=read1();
  for (register int i=1;i<=m;i++) addedge(x[i],y[i],w[i]),addedge(y[i],x[i],w[i]);
  dijkstra(s);
  dijkstran(t);
  init();
  for (register int i=1;i<=m;i++)
  {
     if (dis[t]==dis[x[i]]+w[i]+disn[y[i]])
     {
        insert(x[i],y[i],1);
     }
     if(dis[t]==dis[y[i]]+w[i]+disn[x[i]])
     {
        insert(y[i],x[i],1);
     }
  }
  cout<<dinic();
  return 0;
}

原文地址：https://www.cnblogs.com/yimmortal/p/10160855.html