【牛客网】Whalyzh's Problem

每个\(b_{i,j}\)建一个点，认为选了\(b_{i,j}\)一定会选\(a_{i}\)和\(a_{j}\)

选了\(a_{i}\)的话会带了一个\(-b_{i,i}\)的价值

然后再用01分数规划二分答案，选了\(a_{i}\)还会带来\(-x\)的代价，x是二分的答案

如果正数值减最大流大于0认为这个答案可以达到

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define ba 47
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 +c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}
struct node {int to,next;db cap;
}E[MAXN * 2];
int N,Ncnt,S,T;
int b[105][105],id[105][105];
int head[20005],sumE = 1,cur[20005];
void add(int u,int v,db c) {E[++sumE].to = v;E[sumE].next = head[u];E[sumE].cap = c;head[u] = sumE;
}
void addtwo(int u,int v,db c) {add(u,v,c);add(v,u,0);
}
int dis[20005];
queue<int> Q;
bool BFS() {memset(dis,0,sizeof(dis));while(!Q.empty()) Q.pop();Q.push(S);dis[S] = 1;while(!Q.empty()) {int u = Q.front();Q.pop();for(int i = head[u] ; i ; i = E[i].next) {int v = E[i].to;if(E[i].cap > 1e-6 && !dis[v]) {dis[v] = dis[u] + 1;if(v == T) return true;Q.push(v);}}}return false;
}
db dfs(int u,db aug) {if(u == T) return aug;for(int &i = cur[u] ; i ; i = E[i].next) {int v = E[i].to;if(E[i].cap > 1e-6 && dis[v] == dis[u] + 1) {db t = dfs(v,min(aug,E[i].cap));if(t > 1e-6) {E[i].cap -= t;E[i ^ 1].cap += t;return t;}}}return 0;
}
db Dinic() {db res = 0;while(BFS()) {for(int i = 1 ; i <= Ncnt ; ++i) cur[i] = head[i];while(db d = dfs(S,1e9) && d > 1e-6) res += d;}return res;
}
bool check(db x) {db res = 0;memset(head,0,sizeof(head));sumE = 1;for(int i = 1 ; i <= N ; ++i) {for(int j = 1 ; j <= N ; ++j) {addtwo(S,id[i][j],b[i][j]);res += b[i][j];addtwo(id[i][j],i,1e9);addtwo(id[i][j],j,1e9);}}for(int i = 1 ; i <= N ; ++i) {addtwo(i,T,b[i][i] + x);}return res - Dinic() > 1e-6;
}
void Solve() {read(N);Ncnt = N;for(int i = 1 ; i <= N ; ++i) {for(int j = 1 ; j <= N ; ++j) {read(b[i][j]);id[i][j] = ++Ncnt;}}S = ++Ncnt;T = ++Ncnt;db L = 0.0,R = 1e5;int cnt = 50;while(cnt--) {db mid = (L + R) / 2.0;if(check(mid)) L = mid;else R = mid;}printf("%.5lf\n",L);
}
int main(){
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifint T;read(T);for(int i = 1 ; i <= T ; ++i) Solve();return 0;
}