【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 ivorysi

    freopen("f1.in","r",stdin);

#endif

    int T;

    read(T);

    for(int i = 1 ; i <= T ; ++i) Solve();

    return 0;

}

巴特西

【牛客网】Whalyzh's Problem

【牛客网】Whalyzh's Problem

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