最关键的就是找好所有的要满足的不等式条件，注意隐含的条件还有一点就是注意没有源点 建立源点
#2436. 「SCOI2011」糖果
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int,int>;
#define int long long
const int N = 5e5+10,M = 2e6+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int gcd(int a,int b){return b?a:gcd(b,a%b);}
int lcm(int a,int b){return a*b/gcd(a,b);}
int qmi(int a,int b,int mod){int res=1;while(b){if(b&1)res=res*a%mod;b>>=1;a=a*a%mod;}return res;}int n,q,m;
int dist[N];
bool vis[N];
int cnt[N];
int e[M],ne[M],w[M],h[N],idx;
void add(int a,int b,int c){e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
}void spfa()
{stack<int>q;memset(dist,-0x3f,sizeof dist);dist[0] = 0;vis[0] = true;q.push(0);//int count = 0;while(q.size()){auto t = q.top();q.pop();vis[t] = false;for(int i=h[t];~i;i=ne[i]){int j = e[i];if(dist[j]<dist[t]+w[i]){dist[j] = dist[t]+w[i];cnt[j]++;if(cnt[j]>=n){cout<<-1;return;}//if(++count>100000){cout<<-1;return;}if(!vis[j]){vis[j] = true;q.push(j);}}}}int res = 0;for(int i=1;i<=n;i++)res+=dist[i];cout<<res;}void solve()
{cin>>n>>m;memset(h,-1,sizeof h);for(int i=1;i<=m;i++){int a,b,c;cin>>c>>a>>b;if(c==1){add(a,b,0),add(b,a,0);}else if(c==2){add(a,b,1);}else if(c==3){add(b,a,0);}else if(c==4){add(b,a,1);}else add(a,b,0);}for(int i=1;i<=n;i++)add(0,i,1);spfa();}signed main()
{ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int _;//cin>>_;_ = 1;while(_--)solve();return 0;
}
#10087. 「一本通 3.4 例 1」Intervals
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int,int>;
#define int long long
const int N = 2e5+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int gcd(int a,int b){return b?a:gcd(b,a%b);}
int lcm(int a,int b){return a*b/gcd(a,b);}
int qmi(int a,int b,int mod){int res=1;while(b){if(b&1)res=res*a%mod;b>>=1;a=a*a%mod;}return res;}int n,q,m;int e[N],ne[N],w[N],h[N],idx;
void add(int a,int b,int c){e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
}int dist[N];
bool vis[N];
int mx;
void spfa()
{memset(dist,-0x3f,sizeof dist);queue<int>q;q.push(0);dist[0] = 0;vis[0] = true;while(q.size()){auto t = q.front();q.pop();vis[t] = false;//cout<<t<<"\n";for(int i=h[t];~i;i=ne[i]){int j = e[i];if(dist[j]<dist[t]+w[i]){dist[j] = dist[t]+w[i];if(!vis[j]){vis[j] = true;q.push(j);}}}}cout<<dist[50001];
}void solve()
{memset(h,-1,sizeof h);cin>>m;for(int i=1;i<=50001;++i){add(i-1,i,0),add(i,i-1,-1);}while(m--){int a,b,c;cin>>a>>b>>c;a++,b++;add(a-1,b,c);}spfa();}signed main()
{ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int _;//cin>>_;_ = 1;while(_--)solve();return 0;
}
#10090. 「一本通 3.4 练习 2」布局 Layout
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int,int>;
#define int long long
const int N = 1e5+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int gcd(int a,int b){return b?a:gcd(b,a%b);}
int lcm(int a,int b){return a*b/gcd(a,b);}
int qmi(int a,int b,int mod){int res=1;while(b){if(b&1)res=res*a%mod;b>>=1;a=a*a%mod;}return res;}int n,q,m1,m2;
int dist[N],cnt[N];
bool vis[N];
int e[N],ne[N],w[N],h[N],idx;
void add(int a,int b,int c){e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
}bool spfa(int sz)
{memset(dist,0x3,sizeof dist);memset(cnt,0,sizeof cnt);memset(vis,0,sizeof vis);queue<int>q;for(int i=1;i<=sz;++i){dist[i] = 0;q.push(i);vis[i] = true;}while(q.size()){auto t = q.front();q.pop();vis[t] = false;for(int i=h[t];~i;i=ne[i]){int j = e[i];if(dist[j]>dist[t]+w[i]){dist[j] = dist[t]+w[i];cnt[j]++;if(cnt[j]>=n)return true;if(!vis[j]){vis[j] = true;q.push(j);}}}}return false;}void solve()
{cin>>n>>m1>>m2;memset(h,-1,sizeof h);for(int i=0;i<=n;++i)add(i+1,i,0);while(m1--){int a,b,c;cin>>a>>b>>c;if(a>b)swap(a,b);add(a,b,c);} while(m2--){int a,b,c;cin>>a>>b>>c;if(a>b)swap(a,b);add(b,a,-c);}if(spfa(n))cout<<-1;else{spfa(1);if(dist[n]>inf/2)cout<<-2;else cout<<dist[n];}}signed main()
{ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int _;//cin>>_;_ = 1;while(_--)solve();return 0;
}
