[Cqoi2014]危桥 (两遍网络流)

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题目链接

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 inline int read()
  5 {
  6     int x=0,f=1;char ch=getchar();
  7     while(ch<0||ch>9){if(ch==-)f=-1;ch=getchar();}
  8     while(ch>=0&&ch<=9){x=x*10+ch-0;ch=getchar();}
  9     return x*f;
 10 }
 11 
 12 /********************************************************************/
 13 
 14 #define inf 0xffffff
 15 #define T 2001
 16 const int maxn = 2e6+7;
 17 const int Maxn = 2e3+5;
 18 int a, b;
 19 int ans1, ans2;
 20 int head[Maxn], q[Maxn], dis[Maxn], from[Maxn];
 21 bool vis[Maxn];
 22 
 23 struct node
 24 {
 25     int to, from, Next;
 26     int v, c;    
 27 }e[maxn];
 28 int cnt = 1;
 29 
 30 int gcd(int x, int y){
 31     if(y == 0) return x;
 32     else return gcd(y, x%y);
 33 }
 34 
 35 void add_edge(int u, int v, int w, int c){
 36     e[++cnt].to = v; e[cnt].from = u; e[cnt].Next = head[u]; head[u] = cnt;
 37     e[cnt].v = w; e[cnt].c = c;
 38 }
 39 
 40 void insert(int u, int v, int w, int c){
 41     add_edge(u, v, w, c);
 42     add_edge(v, u, 0, -c);
 43 }
 44 
 45 //是否满足条件
 46 bool check(int x, int y){
 47     if(x < y) swap(x, y);
 48     int t = int(sqrt(x*x-y*y));
 49     return (gcd(y, t) == 1 && x*x-y*y == t*t);
 50 }
 51 
 52 bool spfa(){
 53     for(int i = 0;i <= T;i++){
 54         dis[i] = -inf;
 55     }
 56     int t = 0, w = 1;
 57     dis[0] = 0; q[0] = 0; vis[0] = 1;
 58     while(t != w){
 59         int now = q[t]; t++;
 60         if(t == T) t = 0;
 61         for(int i = head[now];i;i = e[i].Next){
 62             if(e[i].v && e[i].c+dis[now] > dis[e[i].to]){
 63                 dis[e[i].to] = e[i].c + dis[now];
 64                 from[e[i].to] = i;
 65                 if(!vis[e[i].to]){
 66                     vis[e[i].to] = 1;
 67                     q[w++] = e[i].to;
 68                     if(w == T) w = 0;
 69                 }
 70             }
 71         }
 72         vis[now] = 0;
 73     }
 74     if(dis[T] == -inf) return false;
 75     return true;
 76 }
 77 
 78 void dfs(){
 79     int x = inf;
 80     for(int i = from[T];i;i = from[e[i].from]){
 81         x = min(e[i].v, x);
 82     }
 83     for(int i = from[T];i;i = from[e[i].from]){
 84         ans2 += x*e[i].c;
 85         e[i].v -= x;
 86         e[i^1].v += x;
 87     }
 88 }
 89 
 90 int main(){
 91     a = read(); b = read();
 92     for(int i = a;i <= b;i++){
 93         for(int j = a;j <= b;j++){
 94             if(check(i, j) && i != j){
 95                 insert(i, j+1000, 1, i+j);
 96             }
 97         }
 98     }
 99     for(int i = a;i <= b;i++){
100         insert(0, i, 1, 0);
101         insert(i+1000, T, 1, 0);
102     }
103     while(spfa()) dfs();
104     for(int i = 2;i <= cnt;i += 2){
105         if()
106     }
107     return 0;
108 }

 

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