- HDU - 4370
- http://acm.hdu.edu.cn/showproblem.PHP?pid=4370
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:
- X12+X13+...X1n=1
- 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,∑Cij*Xij(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;
- }
- HDU4370
HDU - 4370 0 or 1 最短路
来源: http://www.bubuko.com/infodetail-2787502.html