差分约束

对于

\(a+b>=c\)

若干个这样的不等式

要求判断有无解

可以从最短路中得到启发

\(d[s]+w_{s->v}<=d[v]\)

从\(s\)到\(v\)连一条权值为\(w_{s->v}\)的边

从\(c\)到\(a\)连一条权值为\(-b\)的边

然后 用\(dfs\)优化\(spfa\)

P1993 小K的农场

#include <stack>
#include <queue>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define reg register int
#define isdigit(x) ('0' <= x&&x <= '9')
template<typename T>
inline T Read(T Type)
{
	T x = 0,f = 1;
	char a = getchar();
	while(!isdigit(a)) {if(a == '-') f = -1;a = getchar();}
	while(isdigit(a)) {x = (x << 1) + (x << 3) + (a ^ '0');a = getchar();}
	return x * f;
}
const int MAXN = 10010,MAXM = 20010,inf = 0x3f3f3f;
struct EDGE
{
	int v,w,_nxt;
}edge[MAXM << 1];
int cnt,_ori[MAXN];
inline void add(int u,int v,int w)
{
	edge[++cnt].v = v;
	edge[cnt]._nxt = _ori[u];
	edge[cnt].w = w;
	_ori[u] = cnt; 
}
#define code_out {printf("No");exit(0);}
typedef long long ll;
bool vis[MAXN];
ll d[MAXN];
int n,k;
bool able;
inline void _SPFA(int s)
{
	vis[s] = 1;
	for(reg e = _ori[s],v;e;e = edge[e]._nxt)
	{
		if(d[s] + edge[e].w > d[v = edge[e].v])
		{
			d[v] = d[s] + edge[e].w;
			if(vis[v]) {able = 1;return;}
			_SPFA(v);
			if(able) return;
		}
	}
	vis[s] = 0;
	return;
}
int main()
{
	n = Read(1),k = Read(1);
	for(reg i = 1;i <= k;i++)
	{
		int sit = Read(1),u = Read(1),v = Read(1),w;
		switch(sit)
		{
			case 1:
				w = Read(1);
				if(v == u&&!w) code_out
				add(v,u,w);
				break;
			case 2:
				w = Read(1);
				if(v == u&&!w) code_out
				add(u,v,-w);
				break;
			case 3:
				add(u,v,0),add(v,u,0);
				break;
		}
	}
	for(reg i = 1;i <= n;i++) add(0,i,0),d[i] = -inf;
	_SPFA(0);
	if(!able) printf("Yes");
	else printf("No");
	return 0;
}