2014 西安 The Problem Needs 3D Arrays
题意：给你n个数， 然后1-n的数， 然后要求按顺序选出m个数， 求 逆序数/m 个数的 最大值是多少。
题解：裸的最大密度子图。逆序的2个数建边， 跑一下最大密度子图就AC了。
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define max3(a,b,c) max(a,max(b,c)) 12 #define min3(a,b,c) min(a,min(b,c)) 13 typedef pair<int,int> pll; 14 const LL INF = 0x3f3f3f3f3f3f3f3f; 15 const int inf = 0x3f3f3f3f; 16 const LL mod = (int)1e9+7; 17 const int N = 120; 18 const int M = 500000; 19 double eps = 1e-8; 20 int head[N], deep[N], cur[N]; 21 int u[M], v[M]; 22 int vis[N], d[N]; 23 double w[M]; int to[M], nx[M]; 24 int n, m, tot; 25 int a[N]; 26 void add(int u, int v,double val){ 27 w[tot] = val; to[tot] = v; 28 nx[tot] = head[u]; head[u] = tot++; 29 w[tot] = 0; to[tot] = u; 30 nx[tot] = head[v]; head[v] = tot++; 31 } 32 int bfs(int s, int t){ 33 queue<int> q; 34 memset(deep, 0, sizeof(int)*(n+3)); 35 q.push(s); 36 deep[s] = 1; 37 while(!q.empty()){ 38 int u = q.front(); 39 q.pop(); 40 for(int i = head[u]; ~i; i = nx[i]){ 41 if(w[i] > 0 && deep[to[i]] == 0){ 42 deep[to[i]] = deep[u] + 1; 43 q.push(to[i]); 44 } 45 } 46 } 47 if(deep[t] > 0) return 1; 48 return 0; 49 } 50 double Dfs(int u, int t, double flow){ 51 if(u == t) return flow; 52 for(int &i = cur[u]; ~i; i = nx[i]){ 53 if(deep[u] + 1 == deep[to[i]] && w[i] > 0){ 54 double di = Dfs(to[i], t, min(w[i], flow)); 55 if(di > 0){ 56 w[i] -= di, w[i^1] += di; 57 return di; 58 } 59 } 60 } 61 return 0; 62 } 63 64 int Dinic(int s, int t){ 65 double ans = 0, tmp; 66 while(bfs(s, t)){ 67 for(int i = 0; i <= n+1; i++) cur[i] = head[i]; 68 while(tmp = Dfs(s, t, INF)) ans += tmp; 69 } 70 return ans; 71 } 72 73 bool check(double x){ 74 memset(head, -1, sizeof(int) * (n+2)); 75 tot = 0; 76 int s = 0, t = n + 1; 77 for(int i = 1; i <= m; i++){ 78 add(u[i], v[i], 1); 79 add(v[i], u[i], 1); 80 } 81 for(int i = 1; i <= n; i++){ 82 add(s, i, m); 83 add(i, t, m+2*x-d[i]); 84 } 85 return (m*n-Dinic(s,t))/2.0 >= 1e-6; 86 } 87 88 int main(){ 89 int T; 90 scanf("%d", &T); 91 for(int _i = 1; _i <= T; _i++){ 92 scanf("%d", &n); 93 for(int i = 1; i <= n; i++)scanf("%d", &a[i]); 94 memset(d, 0, sizeof(int)*(n+1)); 95 m = 0; 96 for(int i = 1; i <= n; i++){ 97 for(int j = 1; j < i; j++){ 98 if(a[j] > a[i]){ 99 d[i]++; 100 d[j]++; 101 m++; 102 u[m] = i; 103 v[m] = j; 104 } 105 } 106 } 107 double l = 0, r = m, mid; 108 while(r - l >= eps){ 109 mid = (l+r)/2; 110 if(check(mid)) l = mid; 111 else r = mid; 112 } 113 printf("Case #%d: %.12f\n", _i, l); 114 //printf 115 } 116 117 return 0; 118 }
