題目描述

傳送門

題解

最大流和可行流的做法的區別：先ss->tt做一遍最大流，判斷是否可行；然后將t->s,inf這條邊去掉，再做一遍s->t的最大流，即為答案
這道題原圖的建圖方法是：
對于每一行i，s->i,[a(i,n),a(i,n)+1]
對于每一列j，j->t,[a(n,j),a(n,j)+1]
對于每一個點(i,j)，i->j,[a(i,j),a(i,j)+1]
然后再按照有源匯有上下界對這個圖進行改造即可

代碼

#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<queue> using namespace std; #define N 100005 #define inf 1000000000int n,s,t,ss,tt,maxflow,in,out,ans; double a[105][105]; int l[105][105],r[105][105],p[105][105]; int tot,point[N],nxt[N],v[N],remain[N]; int d[N],deep[N],last[N],cur[N],num[N]; queue <int> q;void addedge(int x,int y,int cap) {++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap;++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0; } void bfs(int t) {for (int i=1;i<=t;++i) deep[i]=t;deep[t]=0;for (int i=1;i<=t;++i) cur[i]=point[i];while (!q.empty()) q.pop();q.push(t);while (!q.empty()){int now=q.front();q.pop();for (int i=point[now];i!=-1;i=nxt[i])if (deep[v[i]]==t&&remain[i^1]){deep[v[i]]=deep[now]+1;q.push(v[i]);}} } int addflow(int s,int t) {int now=t,ans=inf;while (now!=s){ans=min(ans,remain[last[now]]);now=v[last[now]^1];}now=t;while (now!=s){remain[last[now]]-=ans;remain[last[now]^1]+=ans;now=v[last[now]^1];}return ans; } void isap(int s,int t) {bfs(t);for (int i=1;i<=t;++i) ++num[deep[i]];int now=s;while (deep[s]<t){if (now==t){maxflow+=addflow(s,t);now=s;}bool has_find=false;for (int i=cur[now];i!=-1;i=nxt[i])if (deep[v[i]]+1==deep[now]&&remain[i]){has_find=true;cur[now]=i;last[v[i]]=i;now=v[i];break;}if (!has_find){int minn=t-1;for (int i=point[now];i!=-1;i=nxt[i])if (remain[i]) minn=min(minn,deep[v[i]]);if (!(--num[deep[now]])) break;++num[deep[now]=minn+1];cur[now]=point[now];if (now!=s) now=v[last[now]^1];}} } int main() {tot=-1;memset(point,-1,sizeof(point));scanf("%d",&n);for (int i=1;i<=n;++i)for (int j=1;j<=n;++j){scanf("%lf",&a[i][j]);l[i][j]=floor(a[i][j]);r[i][j]=ceil(a[i][j]);}s=n+n+1,t=s+1,ss=t+1,tt=ss+1;for (int i=1;i<n;++i){addedge(s,i,r[i][n]-l[i][n]);d[s]-=l[i][n],d[i]+=l[i][n];addedge(n+i,t,r[n][i]-l[n][i]);d[n+i]-=l[n][i],d[t]+=l[n][i];}for (int i=1;i<n;++i)for (int j=1;j<n;++j){addedge(i,n+j,r[i][j]-l[i][j]);p[i][j]=tot;d[i]-=l[i][j],d[n+j]+=l[i][j];}for (int i=1;i<=t;++i){if (d[i]>0) addedge(ss,i,d[i]),in+=d[i];if (d[i]<0) addedge(i,tt,-d[i]),out-=d[i];}addedge(t,s,inf);if (in!=out) {puts("NO");return 0;}isap(ss,tt);if (maxflow!=in) {puts("NO");return 0;}remain[tot]=remain[tot^1]=0;isap(s,t);for (int i=1;i<n;++i)for (int j=1;j<n;++j)ans+=remain[p[i][j]]+l[i][j];printf("%d\n",ans*3);return 0; }

總結

以上是生活随笔為你收集整理的[BZOJ3698]XWW的难题（有源汇有上下界的最大流）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。