DFS

Depth-first search

By searching to get a tree

Strategy: as long as the point of discovery did not come, it will come. There are more points to go to pick one, if nowhere to go back to back, look there's no point to go through.

Looking for the path on the map [minority] can be used to solve the shortest path: the shortest path can not solve sequential, which is related through this road before ignition if the right side of the path, it can only be found deep

Recursive form to solve the problem

There aftereffect of choice

Combinatorial problem

State may be a lot, so the data is generally small range

1, represents the state

2, pruning

Pruning method:

Best answer pruning

Memory of pruning

Feasibility pruning

……

1, the flood [1s 32M]

answer

Small-scale data, search

2, hard old gardener [1s 128M] [1s 32M]

answer

FIG contravened find groups (simply, a group is complete subgraph of G)

FIG built originally entered can not have even edges, after anti FIG construction, even the sides of the can is communication between the two markers

Pruning:

1. If a point and i do not even current point edge, then after never even side, it can be deleted from the point i largest group of candidates

2. Use feasible point pruning sum + tot <= ans, Discard

3. The answer to the current use of pruning ans (2) + tot <= ans

(Tot is the group continues to accumulate weights, sum sequence which is the weight of everything and the rest of the

If the sum + tot <= ans, discarded)

Probably the largest group of dots in a sequence beginning in

Remove a point, expanding, when the sequence is empty, over extended

Take out 1, then the sequence is updated to point 1 connected

Expanding until the queue is empty, looking for the complete group

---> + sum tot <= ans

Maximum main function groups:

Why look for the largest group instead of the maximum independent set it? ?

3, birthday cake [1s 10M]

answer

4, only the number of target-shaped [2s128M]

answer

5, chessboard [1s 16M]

answer

BFS

1, Room Escape [1s 64M]

answer

queue<pair<int,int> > Q;
int FindPath(pair<int,int> b,pair<int,int> e) {
    for (int i=0;i<n;++i) for (int j=0;j<m;++j) dis[i][j]=1e9+10;
    Q.push(b); dis[b.first][b.second]=0;
    while (!Q.empty()) {
        pair<int,int> u=Q.front(); Q.pop();
        int x=u.first,y=u.second;
        for (int i=0;i<4;++i) {
            int tx=x+dx[i],ty=y+dy[i];
            if (CoordValid(tx,ty) && mp[tx][ty]!=0 && dis[tx][ty]>dis[x][y]+1) {
                dis[tx][ty]=dis[x][y]+1;
                Q.push(make_pair(tx,ty));    
            }
        }
    }
    return dis[e.first][e.second];
}