P3324 [SDOI2015]星际战争

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cmath>
  4 #include<algorithm>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<cctype>
  8 #include<stack>
  9 #include<queue>
 10 #include<vector>
 11 using namespace std;
 12 #define enter printf("\n")
 13 #define space printf(" ")
 14 #define Mem(a) memset(a, 0, sizeof(a));
 15 typedef long long ll;
 16 typedef double db;
 17 const int INF = 0x3f3f3f3f;
 18 const db eps = 1e-8;
 19 const int maxn = 2e5 + 5;
 20 inline ll read()
 21 {
 22     ll ans = 0;
 23     char ch = getchar(), last = ‘ ‘;
 24     while(!isdigit(ch)) {last = ch; ch = getchar();}
 25     while(isdigit(ch))
 26     {
 27         ans = ans * 10 + ch - ‘0‘; ch = getchar();
 28     }
 29     if(last == ‘-‘) ans = -ans;
 30     return ans;
 31 }
 32 inline void write(ll x)
 33 {
 34     if(x < 0) putchar(‘-‘), x = -x;
 35     if(x >= 10) write(x / 10);
 36     putchar(x % 10 + ‘0‘);    
 37 }
 38 //mrclr//
 39 
 40 int n, m, t;
 41 ll a[60], b[60];
 42 bool s[60][60];
 43 struct Edge
 44 {
 45     int from, to;
 46     ll cap, flow;
 47 };
 48 vector<Edge> edges;
 49 vector<int> G[maxn];
 50 
 51 void init()
 52 {
 53     edges.clear();
 54     for(int i = 0; i < maxn; ++i) G[i].clear();
 55     
 56 }
 57 void addEdge(int from, int to, ll w)            //以下就是网络流的板子了 
 58 {
 59     edges.push_back((Edge){from, to, w, 0});
 60     edges.push_back((Edge){to, from, 0, 0});
 61     int sz = edges.size();
 62     G[from].push_back(sz - 2);
 63     G[to].push_back(sz - 1);
 64 }
 65 
 66 int dis[maxn];
 67 bool vis[maxn];
 68 bool bfs()
 69 {
 70     Mem(vis);
 71     queue<int> q;
 72     q.push(0); vis[0] = 1;
 73     dis[0] = 0;
 74     while(!q.empty())
 75     {
 76         int now = q.front(); q.pop();
 77         for(int i = 0; i < (int)G[now].size(); ++i)
 78         {
 79             Edge& e = edges[G[now][i]];
 80             if(!vis[e.to] && e.cap > e.flow)
 81             {
 82                 vis[e.to] = 1;
 83                 dis[e.to] = dis[now] + 1;
 84                 q.push(e.to);
 85             }
 86         }
 87     }
 88     return vis[t];
 89 }
 90 int cur[maxn];
 91 ll dfs(int now, ll a)
 92 {
 93     if(now == t || !a) return a;
 94     ll flow = 0, f;
 95     for(int& i = cur[now]; i < (int)G[now].size(); ++i)
 96     {
 97         Edge& e = edges[G[now][i]];
 98         if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)
 99         {
100             e.flow += f;
101             edges[G[now][i] ^ 1].flow -= f;
102             flow += f; a -= f;
103             if(!a) break;
104         }
105     }
106     return flow;
107 }
108 
109 ll maxflow()
110 {
111     ll flow = 0;
112     while(bfs())
113     {
114         Mem(cur);
115         flow += dfs(0, (ll)INF * INF);
116     }
117     return flow;
118 }
119 
120 ll sum = 0;
121 
122 bool judge(ll tp)
123 {
124     init();
125     for(int i = 1; i <= m; ++i) addEdge(0, i, b[i] * tp);        //把武器和源点建立一条容量为b[i] * time 的边 
126     for(int i = 1; i <= n; ++i) addEdge(i + m, t, a[i]);        //把机器人和汇点建立一条容量为a[i]的边 
127     for(int i = 1; i <= m; ++i)
128         for(int j = 1; j <= n; ++j)
129             if(s[i][j]) addEdge(i, j + m, INF);            
130     //如果武器i能攻击机器人j，就建边，容量无穷，因为武器i有b[i] * time的限制 
131 //    printf("%lld %lld\n", maxflow(), sum);                
132     //调试的时候请不要这么写，因为maxflow()跑完后已经成残量网络了，在上面跑自然不对 
133     return maxflow() == sum;
134 }
135 
136 int main()
137 {
138     n = read(); m = read();            //防御值扩大1e4倍 
139     t = n + m + 1;
140     for(int i = 1; i <= n; ++i) {a[i] = read(); a[i] *= 10000; sum += a[i];}
141     for(int i = 1; i <= m; ++i) b[i] = read();
142     for(int i = 1; i <= m; ++i)
143         for(int j = 1; j <= n; ++j)
144             {int x = read(); s[i][j] = x;}
145     ll L = 0, R = (ll)INF * (ll)INF;
146     while(L < R)
147     {
148         int mid = (L + R) >> 1;
149         if(judge(mid)) R = mid;
150         else L = mid + 1;
151     }
152     printf("%.5lf\n", (double)L / 10000);
153     return 0;
154 }