"POJ 1151" Air Raid

"POJ 1151" Air Raid

Minimum path coverage problem for Dag graphs (no loop directed graphs)

= max Match + (n-2* Max match)
(Maximum match * * For all the points that can be connected
n-2* maximum match to find all independent points plus maximum matching is the minimum required path)

The following references blog: http://blog.csdn.net/whosemario/article/details/8513836

Defined:

一个有向无环图，要求用尽量少的不相交的简单路径覆盖所有的节点。

Composition:

建立一个二分图，把原图中的所有节点分成两份（X集合为i，Y集合为i‘），如果原来图中有i->j的有向边，则在二分图中建立i->j‘的有向边。最终|最小路径覆盖|=|V|-|M|

Prove:

, the left side of the DAG is constructed with the right dichotomy, and you can find a maximum matching m:1->3 ' of the binary graph.
3->4 ', then how do these two matching edges of M-weight correspond to the edges of the paths in the Dag?
Make an edge of the binary map correspond to one of the forward edges in the Dag: 1->3 ' corresponds to the 1->3 of the left figure, so that 1 in the DAG has a successor node (the only successor of 3 to 1, because a vertex in the binary graph is associated with an edge!) ), so 1 does not become the end vertex in a path in the DAG, and similarly, 3->4 ' 3->4,3 that corresponds to the left figure will not be the end vertex, so there are 4 vertices in total, minus 2 with successive vertices, and two vertices, the end vertex of the DAG path, Each vertex corresponds to a path. Finding the maximum match m in the binary graph is to find the maximum number of non-path end vertices in the corresponding DAG, then in the Dag | v|-| M| is the minimum number of end vertices in the DAG, which is the minimum number of paths for the DAG.

The code is as follows:

#include <cstdlib>#include <cstdio>#include <cstring>using namespace STD;intNBOOLvis[121],mp[121][121];intlink[121];BOOLCanintP) {intI for(i =1; I <= N; ++i) {if(!vis[i] && mp[p][i]) {Vis[i] =true;if(Link[i] = =-1|| Can (Link[i])) {link[i] = p;return 1; } Vis[i] =false; }    }return 0;}intMain () {intt,m,a,b,cnt;scanf("%d", &t); while(t--) {memset(MP,false,sizeof(MP));scanf("%d%d", &n,&m); while(m--) {scanf("%d%d", &a,&b); MP[A][B] =true; }memset(link,-1,sizeof(link)); CNT =0; for(inti =1; I <= N; ++i) {memset(Vis,0,sizeof(VIS));if(Can (i)) cnt++; }printf("%d\n", n-cnt); }return 0;}

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"POJ 1151" Air Raid