HDU - 4370 0 or 1 最短路

参考：https://www.cnblogs.com/hollowstory/p/5670128.html

题意：

　　给定一个矩阵C，构造一个A矩阵，满足条件：

　　　　1.X₁₂+X₁₃+...X_1n=1
　　　　2.X_1n+X_2n+...X_n-1n=1
　　　　3.for each i (1<i<n), satisfies ∑X_ki (1<=k<=n)=∑X_ij (1<=j<=n).

　　使得∑C_ij*X_ij(1<=i,j<=n)最小。

思路：

　　理解条件之前先转换一下思维，将矩阵C看做描述N个点花费的邻接矩阵

　　再来看三个条件：

　　　　条件一：表示1号点出度为1

　　　　条件二：表示n号点入度为1

　　　　条件三：表示k( 1 < k < n )号点出度等于入度

　　最后再来看看题目要求，∑C_ij*X_ij(1<=i,j<=n)，很明显，这是某个路径的花费，而路径的含义可以有以下两种：

　　一：1号点到n号点的花费

　　二：1号点经过其它点成环，n号点经过其它点成环，这两个环的花费之和

　　于是，就变成了一道简单的最短路问题

　　关于环花费的算法，可以改进spfa算法，初始化dis[start] = INF，且一开始让源点之外的点入队

#include <algorithm>

#include  <iterator>

#include  <iostream>

#include   <cstring>

#include   <cstdlib>

#include   <iomanip>

#include    <bitset>

#include    <cctype>

#include    <cstdio>

#include    <string>

#include    <vector>

#include     <stack>

#include     <cmath>

#include     <queue>

#include      <list>

#include       <map>

#include       <set>

#include   <cassert>

using namespace std;

//#pragma GCC optimize(3)

//#pragma comment(linker, "/STACK:102400000,102400000")  //c++

// #pragma GCC diagnostic error "-std=c++11"

// #pragma comment(linker, "/stack:200000000")

// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#define lson (l , mid , rt << 1)

#define rson (mid + 1 , r , rt << 1 | 1)

#define debug(x) cerr << #x << " = " << x << "\n";

#define pb push_back

#define pq priority_queue

typedef long long ll;

typedef unsigned long long ull;

typedef pair<ll ,ll > pll;

typedef pair<int ,int > pii;

typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q

//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q

#define fi first

#define se second

//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)

#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行

#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)

#define max3(a,b,c) max(max(a,b), c);

//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //18

// const int mod = 10007;

const double esp = 1e-;

const double PI=acos(-1.0);

const double PHI=0.61803399;    //黄金分割点

const double tPHI=0.38196601;

template<typename T>

inline T read(T&x){

    x=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

/*-----------------------showtime----------------------*/

            const int maxn = ;

            int n;

            int dis[maxn],a[maxn][maxn],vis[maxn];

            void spfa(int s){

                stack<int>q;

                for(int i=; i<=n; i++){

                    dis[i] = a[s][i];

                    if(i!=s){

                        q.push(i);

                        vis[i] = true;

                    }

                    else vis[i] = false;

                }

                dis[s] = inf;

                while(!q.empty()){

                    int u = q.top();q.pop();

                    vis[u] = false;

                    for(int i=; i<=n; i++){

                        if(u==i)continue;

                        if(dis[i] > dis[u] + a[u][i]){

                            dis[i] = dis[u] + a[u][i];

                            if(vis[i] == false)q.push(i), vis[i] = true;

                        }

                    }

                }

            }

int main(){

            while(~scanf("%d", &n)){

                for(int i=; i<=n; i++){

                    for(int j=; j<=n; j++){

                        scanf("%d", &a[i][j]);

                    }

                }

                spfa();

                int ans = dis[n];

                int a1 = dis[];

                spfa(n);

                a1 += dis[n];

                printf("%d\n", min(a1, ans));

            }

            return ;

}

HDU4370

巴特西

HDU - 4370 0 or 1 最短路

题意：

思路：

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