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。。。
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <cmath> 6 #include <string> 7 #include <vector> 8 #include <queue> 9 #include <set> 10 using namespace std; 11 12 #define ll long long 13 #define inf 2100000000 14 #define eps 1e-8 15 #define mn 505 16 #define me 200005 17 18 int dis[mn], pre[mn], gap[mn], arc[mn], f[me], cap[me]; 19 int first[mn], nxt[me], vv[me], e; 20 21 inline void add(int u,int v,int c) { 22 vv[e] = v, cap[e] = c, nxt[e] = first[u], first[u] = e++; 23 vv[e] = u, cap[e] = 0, nxt[e] = first[v], first[v] = e++; 24 } 25 int sap( int s, int t, int n ) { 26 int q[mn], j, mindis, ans = 0, head = 0, tail = 1, u, v, low; 27 bool found, vis[mn]; 28 memset( dis, 0, sizeof(dis) ); 29 memset( gap, 0, sizeof(gap) ); 30 memset( vis, 0, sizeof(vis) ); 31 memset( arc, -1, sizeof(arc) ); 32 memset( f, 0, sizeof(f) ); 33 q[0] = t; vis[t] = true; dis[t] = 0; gap[0] = 1; 34 while( head < tail ) { 35 u = q[head++]; 36 for( int i = first[u]; i != -1; i = nxt[i] ) { 37 v = vv[i]; 38 if( !vis[v] ) { 39 dis[v] = dis[u] + 1; 40 vis[v] = true; 41 q[tail++] = v; 42 gap[dis[v]]++; 43 arc[v] = first[v]; 44 } 45 } 46 } 47 u = s; low = inf; pre[s] = s; 48 while( dis[s] < n ) { 49 found = false; 50 for( int &i = arc[u]; i != -1; i = nxt[i] ) 51 if( dis[vv[i]] == dis[u]-1 && cap[i] > f[i] ) { 52 found = true; v = vv[i]; 53 low = low < cap[i] - f[i] ? low : cap[i] - f[i]; 54 pre[v] = u; u = v; 55 if( u == t ) { 56 while( u != s ) { 57 u = pre[u]; 58 f[arc[u]] += low; 59 f[arc[u]^1] -= low; 60 } 61 ans += low; low = inf; 62 } 63 break; 64 } 65 if( found ) 66 continue; 67 mindis = n; 68 for(int i = first[u]; i != -1; i = nxt[i] ) { 69 if( mindis > dis[vv[i]] && cap[i] > f[i] ) { 70 mindis = dis[vv[j = i]]; 71 arc[u] = i; 72 } 73 } 74 gap[dis[u]]--; 75 if( gap[dis[u]] == 0 ) 76 return ans; 77 dis[u] = mindis + 1; 78 gap[dis[u]]++; 79 u = pre[u]; 80 } 81 return ans; 82 } 83 84 char ch[111][111]; 85 bool border[mn]; 86 int a[111]; 87 int getborder(int n){ 88 memset(border,false,sizeof(border)); 89 for(int i=1;i<=n;++i){ 90 if(a[i]==0) 91 for(int j=1;j<=n;++j) 92 if(a[j]&&ch[i][j]=='Y')border[j]=true; 93 } 94 int ret=0; 95 for(int i=1;i<=n;++i)ret+=border[i]; 96 return ret; 97 } 98 void build(int cap,int n){ 99 memset(first,-1,sizeof(first));e=0; 100 for(int i=1;i<=n;++i)if(a[i])add(2*n+1,i,a[i]-1),add(i,i+n,a[i]); 101 for(int i=1;i<=n;++i) 102 for(int j=i+1;j<=n;++j) 103 if(ch[i][j]=='Y'&&a[i]&&a[j]) 104 add(i+n,j,inf),add(j+n,i,inf); 105 for(int i=1;i<=n;++i)if(border[i])add(i,2*n+2,cap); 106 } 107 int main(){ 108 int t; 109 scanf("%d",&t); 110 while(t--){ 111 int n; 112 scanf("%d",&n); 113 for(int i=1;i<=n;++i)scanf("%d",a+i); 114 for(int i=1;i<=n;++i)scanf("%s",ch[i]+1); 115 int cnt=getborder(n); 116 int l=0,r=10000; 117 while(l<r){ 118 int mid=(l+r+1)/2; 119 build(mid,n); 120 if(sap(2*n+1,2*n+2,2*n+2)!=cnt*mid)r=mid-1; 121 else l=mid; 122 } 123 printf("%d ",l+1); 124 } 125 return 0; 126 }