HDU 1853 & HDU 3488【有向環最小權值覆蓋問題 】帶權二分圖匹配 KM算法

In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000) one-way roads connecting them. You are lucky enough to have a chance to have a tour in the kingdom. The route should be designed as: The route should contain one or more loops. (A loop is a route like: A->B->……->P->A.)?
Every city should be just in one route.?
A loop should have at least two cities. In one route, each city should be visited just once. (The only exception is that the first and the last city should be the same and this city is visited twice.)?
The total distance the N roads you have chosen should be minimized.?

Input

An integer T in the first line indicates the number of the test cases.?
In each test case, the first line contains two integers N and M, indicating the number of the cities and the one-way roads. Then M lines followed, each line has three integers U, V and W (0 < W <= 10000), indicating that there is a road from U to V, with the distance of W.?
It is guaranteed that at least one valid arrangement of the tour is existed.?
A blank line is followed after each test case.

Output

For each test case, output a line with exactly one integer, which is the minimum total distance.

Sample Input

1 6 9 1 2 5 2 3 5 3 1 10 3 4 12 4 1 8 4 6 11 5 4 7 5 6 9 6 5 4

Sample Output

42

題意：

給出n個點m條單向邊邊以及經過每條邊的費用，讓你求出走過一個哈密頓環（除起點外，每個點只能走一次）的最小費用。

解析：

任意類似的【有向環最小權值覆蓋】問題，都可以用最小費用流來寫。

由于題目中要求每個點最多走一次，為了防止走多次的發生，我們要把每個點 i 拆成左點i 和 右點i + n兩個點。

具體建圖如下:

源點outset編號0, 所有左點編號 1~n ，右點編號?n+1 ~ 2*n, 匯點inset 編號 2*n+1.
（1）源點outset到第i個點有邊 ( outset, i, 1, 0)，即源點和左點建邊。
（2）如果從 i 點到 j 點有權值為 c 的邊,那么有邊 ?(i, ?j+n, ?1, ?c)，即左點和右點建邊， 確保了每個點只走一次。
（3）每個節點到匯點有邊 (i+n, ?inset, ?1, ?0)， 即右點和匯點建邊。

最終如果最大流 == n 的話（即滿流），那么最小費用就是我們所求；若最大流量小于n，則不存在滿足條件的環

為什么這樣的構圖方法就可以求得我們所要的解， 具體解析請點這里：解析

而且本題時間要求比較嚴格，而且點少邊多；各種超時，需要加個去重邊處理才行。即先用鄰接矩陣存儲最小邊權，然后再建圖

#include<stdio.h> #include<iostream> #include<string.h> #include<queue> #include<cstdio> #include<string> #include<math.h> #include<algorithm> #include<map> #include<set> #include<stack> #define mod 998244353 #define INF 0x3f3f3f3f #define eps 1e-6 using namespace std; typedef long long ll; #define MAXN 1003 #define MAXM 40004 //最小費用最大流 struct Edge{int to,next;int flow,cost,cap; }edge[MAXM]; int tol,head[MAXN]; void init() {tol=0;memset(head,-1,sizeof head); } void addEdge(int u,int v,int cap,int cost){edge[tol].to=v;edge[tol].cap=cap;edge[tol].cost=cost;edge[tol].flow=0;edge[tol].next=head[u];head[u]=tol++;edge[tol].to=u;edge[tol].cap=0;edge[tol].cost=-cost;edge[tol].flow=0;edge[tol].next=head[v];head[v]=tol++; }bool inq[MAXN];//標記是否點是否在隊列 int dis[MAXN];//最短距離 int pre[MAXN];//記錄路徑 int q[MAXN*10];//隊列 //單位費用可能是負值，所以用SPFA bool spfa(int st,int en) {memset(inq,0,sizeof inq);memset(dis,INF,sizeof dis);memset(pre,-1,sizeof pre);int rear=0,front=0;dis[st]=0;inq[st]=true;q[front++]=st;while(rear<front){int u=q[rear++];inq[u]=false;for(int e=head[u];e!=-1;e=edge[e].next){int v=edge[e].to;if(edge[e].cap>edge[e].flow&&dis[v]>dis[u]+edge[e].cost){dis[v]=dis[u]+edge[e].cost;pre[v]=e;//表示邊e-->v,e就是v的前驅if(!inq[v])inq[v]=true,q[front++]=v;}}}return pre[en]!=-1; } int MCMF(int st,int en,int &cost,int &flow) {//如果能找到從源點到匯點的最短路，說明還沒有達到最小費用最大流while(spfa(st,en)){int Min=INF;//最小殘余流量//沿著當前路徑返回for(int i=pre[en];i!=-1;i=pre[edge[i^1].to]){int rem=edge[i].cap-edge[i].flow;Min=Min>rem?rem:Min;}for(int i=pre[en];i!=-1;i=pre[edge[i^1].to]){edge[i].flow+=Min;//正向邊添加殘余流量edge[i^1].flow-=Min;//反向邊減少殘余流量cost+=Min*edge[i].cost;}flow+=Min;} } //以上為最小費用最大流模板 int mp[210][210]; int main() {int n,m;int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);memset(mp,INF,sizeof mp);init();int st=0,en=2*n+1;for(int i=1;i<=n;i++){addEdge(st,i,1,0);addEdge(n+i,en,1,0);}//去重邊操作for(int i=1;i<=m;i++){int u,v,w;scanf("%d%d%d",&u,&v,&w);mp[u][v]=min(mp[u][v],w);//addEdge(u,n+v,1,w);}//建圖for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(mp[i][j]!=INF)addEdge(i,j+n,1,mp[i][j]);}}int cost=0,flow=0;MCMF(st,en,cost,flow);//printf("%d %d\n",flow,cost);if(flow==n)printf("%d\n",cost);elseprintf("-1\n");} }

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