题意：自己去看吧,最小化每行每列所有元素与给定矩阵差的和的绝对值中的最大值
又带绝对值又带max不方便直接求
显然可以二分这个最大值 然后判定问题，给定矩阵每行每列的范围和每个元素的取值范围判断可行...和之前做过的一样了上下界可行流 1A好开心啊#include#include #include #include using namespace std;#define fir first#define sec secondtypedef long long ll;const int N=2005, M=1e5+5, INF=1e9;inline ll read(){ char c=getchar();ll x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f;}int n, m, l, r, a, row[N], col[N], s, t, extra[N], tot;struct edge{int v, c, f, ne;}e[M];int cnt=1, h[N];inline void ins(int u, int v, int c) { e[++cnt]=(edge){v, c, 0, h[u]}; h[u]=cnt; e[++cnt]=(edge){u, 0, 0, h[v]}; h[v]=cnt;}int q[N], head, tail, vis[N], d[N], cur[N];bool bfs(int s, int t) { memset(vis, 0, sizeof(vis)); head=tail=1; q[tail++]=s; d[s]=0; vis[s]=1; while(head!=tail) { int u=q[head++]; for(int i=h[u];i;i=e[i].ne) if(!vis[e[i].v] && e[i].c>e[i].f) { vis[e[i].v]=1; d[e[i].v]=d[u]+1; q[tail++]=e[i].v; if(e[i].v == t) return true; } } return false;}int dfs(int u, int a, int t) { if(u==t || a==0) return a; int flow=0, f; for(int &i=cur[u];i;i=e[i].ne) if(d[e[i].v]==d[u]+1 && (f=dfs(e[i].v, min(a, e[i].c-e[i].f), t))>0) { flow+=f; e[i].f+=f; e[i^1].f-=f; a-=f; if(a==0) break; } if(a) d[u]=-1; return flow;}int dinic(int s, int t) { int flow=0; while(bfs(s, t)) { for(int i=0; i<=tot; i++) cur[i]=h[i]; flow+=dfs(s, INF, t); } return flow;}bool check(int mid) { //printf("\ncheck %d\n",mid); s=0; t=n+m+1; cnt=1; memset(h,0,sizeof(h)); memset(extra,0,sizeof(extra)); for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) ins(i, n+j, r-l), extra[i]-=l, extra[n+j]+=l; for(int i=1; i<=n; i++) { int l = -mid+row[i], r = mid+row[i]; ins(s, i, r-l), extra[s]-=l, extra[i]+=l; } for(int j=1; j<=m; j++) { int l = -mid+col[j], r = mid+col[j]; ins(n+j, t, r-l), extra[n+j]-=l, extra[t]+=l; } int ss=t+1, tt=t+2, sum=0; tot=tt; for(int i=s; i<=t; i++) { if(extra[i]>0) ins(ss, i, extra[i]), sum+=extra[i]; if(extra[i]<0) ins(i, tt, -extra[i]); } ins(t, s, INF); int flow=dinic(ss, tt); //printf("flow %d %d\n",flow,sum); return flow==sum;}void solve() { int l=0, r=200000, ans=-1; while(l<=r) { int mid=(l+r)>>1; if(check(mid)) ans=mid, r=mid-1; else l=mid+1; } printf("%d",ans);}int main() { freopen("in","r",stdin); n=read(); m=read(); for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) a=read(), row[i]+=a, col[j]+=a; l=read(); r=read(); solve();}