[SDOI2013]费用流

嘟嘟嘟

看到这题目如果想费用流的话就gg了，就像LCA这道题一样，不要被题目迷惑……

我们考虑Bob，他想让总费用最大，那么他的所有费用一定都加到流量最大的那一条边上。而Alice想让总费用最小，那么就是让流量最大的那一条边最小，自然就想到二分边的容量啦！

不过这道题特殊的地方是要实数二分。其实终止条件就是当R - L小于自己设定的一个精度。然后如果当前不符合，不应该是L = mid + 1,因为(mid, mid + 1)中很有可能是答案，所以应该是L = mid +eps。

第一次写实数二分，总是死循环，调了有那么一会儿……

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cmath>
  4 #include<algorithm>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<cctype>
  8 #include<vector>
  9 #include<stack>
 10 #include<queue>
 11 using namespace std;
 12 #define enter puts("") 
 13 #define space putchar(' ')
 14 #define Mem(a, x) memset(a, x, sizeof(a))
 15 #define rg register
 16 typedef long long ll;
 17 typedef double db;
 18 const int INF = 0x3f3f3f3f;
 19 const db eps = 1e-8;
 20 const int maxn = 105;
 21 inline ll read()
 22 {
 23     ll ans = 0;
 24     char ch = getchar(), last = ' ';
 25     while(!isdigit(ch)) {last = ch; ch = getchar();}
 26     while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
 27     if(last == '-') ans = -ans;
 28     return ans;
 29 }
 30 inline void write(ll x)
 31 {
 32     if(x < 0) x = -x, putchar('-');
 33     if(x >= 10) write(x / 10);
 34     putchar(x % 10 + '0');
 35 }
 36 
 37 int n, m, p, Max;
 38 
 39 struct Edge
 40 {
 41     int from, to;
 42     db cap, flow;
 43 };
 44 vector<Edge> edges;
 45 vector<int> G[maxn];
 46 void addEdge(int from, int to, int w)
 47 {
 48     edges.push_back((Edge){from, to, (db)w, 0});
 49     edges.push_back((Edge){to, from, 0, 0});
 50     int sz = edges.size();
 51     G[from].push_back(sz - 2);
 52     G[to].push_back(sz - 1);
 53 }
 54 
 55 int dis[maxn];
 56 bool bfs()
 57 {
 58     Mem(dis, 0); dis[1] = 1;
 59     queue<int> q; q.push(1);
 60     while(!q.empty())
 61     {
 62         int now = q.front(); q.pop();
 63         for(int i = 0; i < (int)G[now].size(); ++i)
 64         {
 65             Edge& e = edges[G[now][i]];
 66             if(!dis[e.to] && e.cap > e.flow + eps)
 67             {
 68                 dis[e.to] = dis[now] + 1;
 69                 q.push(e.to);    
 70             }
 71         }
 72     }
 73     return dis[n];
 74 }
 75 int cur[maxn];
 76 db dfs(int now, db res)
 77 {
 78     if(now == n || res == 0) return res;
 79     db flow = 0, f;
 80     for(int& i = cur[now]; i < (int)G[now].size(); ++i)
 81     {
 82         Edge& e = edges[G[now][i]];
 83         if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(res, e.cap - e.flow))) > 0)
 84         {
 85             e.flow += f;
 86             edges[G[now][i] ^ 1].flow -= f;
 87             flow += f; res -= f;
 88             if(res == 0) break;
 89         }
 90     }
 91     return flow;
 92 }
 93 
 94 db maxflow()
 95 {
 96     db flow = 0;
 97     while(bfs())
 98     {
 99         Mem(cur, 0);
100         flow += dfs(1, (db)INF);
101     }
102     return flow;
103 }
104 
105 void build_Gra(db x)        //不用重建，修改每一条边的参数就行 
106 {
107     for(int i = 0; i < (int)edges.size(); i += 2)
108     {
109         edges[i].cap = x; edges[i].flow = 0;
110         edges[i ^ 1].cap = 0; edges[i ^ 1].flow = 0;
111     }
112 }
113 
114 int main()
115 {
116     n = read(); m = read(); p = read();
117     for(int i = 1; i <= m; ++i)
118     {
119         int x = read(), y = read(), w = read();
120         addEdge(x, y, w);
121     }
122     Max = maxflow();
123     db L = 0, R = (db)INF;
124     while(L < R - eps)
125     {
126         db mid = (L + R) / 2.00;
127         build_Gra(mid);
128         if(maxflow() > (db)Max - eps) R = mid;
129         else L = mid + eps;
130     }
131     printf("%d %.6lf\n", Max, (L + R) / 2.00 * p);
132     return 0;
133 }

View Code