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+...X1n=1
2.X1n+X2n+...Xn-1n=1
3.for each i (1<i<n), satisfies ∑Xki (1<=k<=n)=∑Xij (1<=j<=n).
使得∑Cij*Xij(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; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 // const int mod = 10007; const double esp = 1e-8; 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=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 309; int n; int dis[maxn],a[maxn][maxn],vis[maxn]; void spfa(int s){ stack<int>q; for(int i=1; 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=1; 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=1; i<=n; i++){ for(int j=1; j<=n; j++){ scanf("%d", &a[i][j]); } } spfa(1); int ans = dis[n]; int a1 = dis[1]; spfa(n); a1 += dis[n]; printf("%d\n", min(a1, ans)); } return 0; }