UVALive 4949 Risk（二分网络流、SAP）

n个区域，每个区域有我方军队a[i]，a[i]==0的区域表示敌方区域，输入邻接矩阵。问经过一次调兵，使得我方边界处（与敌军区域邻接的区域）士兵的最小值最大。输出该最大值。调兵从i->j仅当a[i]>0&&a[j]>0&&adj[i][j]==true;感觉有点像玩三国志什么的。。。

赛后才知道是网络流。。网络流的构图真妙。。。给我方建个超级基地，然后把敌方的区域合并成汇点。。

从超级基地连一条边到我方所有区域，流量为a[i]-1，限流该区域的答案，然后i->i+n，流量为a[i]，用于可能的分配士兵，还有就是i+n->j（a[i]&&a[j]&&adj[i][j]），流量为inf。。。然后再把所有我方的边界区域连到汇点i->T（isBorder[i]）。。。这个流量应该就是答案了。。。

所有正解就是，二分流量答案，SAP检查之。。。

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <set>
 using namespace std;
 
 #define ll long long
 #define inf 2100000000
 #define eps 1e-8
 #define mn 505
 #define me 200005
 
 int dis[mn], pre[mn], gap[mn], arc[mn], f[me], cap[me];
 int first[mn], nxt[me], vv[me], e;
 
 inline void add(int u,int v,int c) {
     vv[e] = v, cap[e] = c, nxt[e] = first[u], first[u] = e++;
     vv[e] = u, cap[e] = , nxt[e] = first[v], first[v] = e++;
 }
 int sap( int s, int t, int n ) {
     int q[mn], j, mindis, ans = , head = , tail = , u, v, low;
     bool found, vis[mn];
     memset( dis, , sizeof(dis) );
     memset( gap, , sizeof(gap) );
     memset( vis, , sizeof(vis) );
     memset( arc, -, sizeof(arc) );
     memset( f, , sizeof(f) );
     q[] = t; vis[t] = true; dis[t] = ; gap[] = ;
     while( head < tail ) {
         u = q[head++];
         for( int i = first[u]; i != -; i = nxt[i] ) {
             v = vv[i];
             if( !vis[v] ) {
                 dis[v] = dis[u] + ;
                 vis[v] = true;
                 q[tail++] = v;
                 gap[dis[v]]++;
                 arc[v] = first[v];
             }
         }
     }
     u = s; low = inf; pre[s] = s;
     while( dis[s] < n ) {
         found = false;
         for( int &i = arc[u]; i != -; i = nxt[i] )
             if( dis[vv[i]] == dis[u]- && cap[i] > f[i] ) {
                 found = true; v = vv[i];
                 low = low < cap[i] - f[i] ? low : cap[i] - f[i];
                 pre[v] = u; u = v;
                 if( u == t ) {
                     while( u != s ) {
                         u = pre[u];
                         f[arc[u]] += low;
                         f[arc[u]^] -= low;
                     }
                     ans += low; low = inf;
                 }
                 break;
             }
         if( found )
             continue;
         mindis = n;
         for(int i = first[u]; i != -; i = nxt[i] ) {
             if( mindis > dis[vv[i]] && cap[i] > f[i] ) {
                 mindis = dis[vv[j = i]];
                 arc[u] = i;
             }
         }
         gap[dis[u]]--;
         if( gap[dis[u]] ==  )
             return ans;
         dis[u] = mindis + ;
         gap[dis[u]]++;
         u = pre[u];
     }
     return ans;
 }
 
 char ch[][];
 bool border[mn];
 int a[];
 int getborder(int n){
     memset(border,false,sizeof(border));
     for(int i=;i<=n;++i){
         if(a[i]==)
             for(int j=;j<=n;++j)
                 if(a[j]&&ch[i][j]=='Y')border[j]=true;
     }
     int ret=;
     for(int i=;i<=n;++i)ret+=border[i];
     return ret;
 }
 void build(int cap,int n){
     memset(first,-,sizeof(first));e=;
     for(int i=;i<=n;++i)if(a[i])add(*n+,i,a[i]-),add(i,i+n,a[i]);
     for(int i=;i<=n;++i)
         for(int j=i+;j<=n;++j)
             if(ch[i][j]=='Y'&&a[i]&&a[j])
                 add(i+n,j,inf),add(j+n,i,inf);
     for(int i=;i<=n;++i)if(border[i])add(i,*n+,cap);
 }
 int main(){
     int t;
     scanf("%d",&t);
     while(t--){
         int n;
         scanf("%d",&n);
         for(int i=;i<=n;++i)scanf("%d",a+i);
         for(int i=;i<=n;++i)scanf("%s",ch[i]+);
         int cnt=getborder(n);
         int l=,r=;
         while(l<r){
             int mid=(l+r+)/;
             build(mid,n);
             if(sap(*n+,*n+,*n+)!=cnt*mid)r=mid-;
             else l=mid;
         }
         printf("%d\n",l+);
     }
     return ;
 }