很容易发现约束条件：L<=K[i,j]*A[i]/B[j]<=U
妈呀这可是乘法啊。。。看起来貌似没法化简。。。
那么看成对数呢？
lg(L)<=lg(K[i,j])+lg(A[i])-lg(B[j])<=lg(U)
这样子就能左移右移了吧=v=
判断是否有负权回路即可。。。
#include#include #include #include #include #include #include #include #define rep(i, l, r) for(int i=l; i<=r; i++)#define down(i, l, r) for(int i=l; i>=r; i--)#define N 456#define MAX 1<<30using namespace std;int read(){ int x=0, f=1; char ch=getchar(); while (ch<'0' || ch>'9') { if (ch=='-') f=-1; ch=getchar(); } while (ch>='0' && ch<='9') { x=x*10+ch-'0'; ch=getchar(); } return x*f;}struct node{int y, n; double v;} e[N*N*4]; int fir[N*2], en;int n, m, l, u, k[N][N], c[N*2];double d[N*2];bool b[N*2], ans;void Add(int x, int y, double v) { en++; e[en].y=y, e[en].v=v, e[en].n=fir[x], fir[x]=en; }int main(){ while (scanf("%d %d %d %d", &n, &m, &l, &u) != EOF) { ans=1; en=0; rep(i, 1, N*2-1) fir[i]=0; rep(i, 1, n) rep(j, 1, m) k[i][j]=read(); rep(i, 1, n) rep(j, 1, m) { Add(i, j+N, log(k[i][j])-log(l)); Add(j+N, i, log(u)-log(k[i][j])); } deque q; rep(i, 1, n) b[i]=1, c[i]=1, d[i]=0, q.push_back(i); rep(i, 1+N, m+N) b[i]=1, c[i]=1, d[i]=0, q.push_back(i); while (!q.empty()) { int x=q.front(), o=fir[x], y=e[o].y; b[x]=0; q.pop_front(); if (c[x] > n) { ans=0; break; } while (o) { if (d[y] > d[x]+e[o].v) { d[y] = d[x]+e[o].v; if (!b[y]) b[y]=1, c[y]++, !q.empty()&&d[y]<=d[q.front()] ? q.push_front(y) : q.push_back(y); } o=e[o].n, y=e[o].y; } } if (ans) printf("YES\n"); else printf("NO\n"); } return 0;}