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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。。。

  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\n",l+1);
124     }
125     return 0;
126 }
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