求边数最小且为奇数条边的环

hdoj 1689 Alien’s Necklace

 枚举每一条边进行dfs。

剪枝：

1.搜索时记录每个点的搜索深度，如果现在的深度小于所记录的深度，才能访问这个点。

2.当出现最小的3时，就不用继续搜索了。

#include <iostream>
#include <vector>
using namespace std;

struct Node {
    int a, b;
}ns[20005];

int n, m, tm[1002], cnt, ans, ca=1, idx;
int mp[1002][1002], from, to;
vector<int> vec[1002];


void input() {
    int i, j, a, b;
    scanf( "%d %d", &n, &m );
    for( i=0; i<=n; ++i ) vec[i].clear();
    for( i=0; i<=n; ++i ) {
        tm[i] = 0;
        for( j=0; j<=n; ++j ) {
            mp[i][j] = 0; 
        }
    }

    idx = 0;
    for( i=0; i<m; ++i ) {
        scanf( "%d %d", &a, &b );
        if( mp[a][b] ) continue;
        ns[idx].a = a;
        ns[idx].b = b;
        idx ++;
        vec[a].push_back(b);
        vec[b].push_back(a);
        mp[a][b] = mp[b][a] = 1;
    }
}



void dfs( int u, int deep ) {
    int i, j, sz;
    if( mp[u][to] ) {
        ++ deep;
        if( (deep&1) && deep < ans ) {
            ans = deep;
        }
        return;
    }

    if( ans == 3 || deep >= ans) return;
    if( deep >= tm[u] && tm[u] ) return;

    tm[u] = deep;
    sz = vec[u].size();
    for( i=0; i<sz; ++i ) dfs( vec[u][i], deep+1 );
}

        

void solve() {
    int i, j, a, b;
    ans = 1000000;
    for( i=0; i<idx; ++i ) {
        from = ns[i].a;
        to = ns[i].b;
        mp[from][to] = mp[to][from] = 0;
        memset( tm, 0, sizeof(tm) );
        dfs( from, 1 );
        if( ans == 3 ) break;
        mp[from][to] = mp[to][from] = 1;
    }

    if( ans == 1000000 ) printf( "Case %d: Poor JYY.\n", ca++ ); 
    else printf( "Case %d: JYY has to use %d balls.\n", ca++, ans );
}    


int main() {
//    freopen( "c:/aaa.txt", "r", stdin );
    int T;
    scanf( "%d", &T );
    while( T-- ) {
        input();
        solve();
    }
    return 0;
}

差不多的一题：

2.hdoj 1599 find the mincost route

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

struct Node {
    int a, b;
}ns[1002];

int mp[102][102], n, m, ans, cnt;
vector<int> vec[102];


void input() {
    int i, j, a, b, c;
    for( i=0; i<=n; ++i ) vec[i].clear();
    for( i=0; i<=n; ++i ) {
        for( j=0; j<=n; ++j ) {
            mp[i][j] = 1000000;
        }
    }
    
    cnt = 0;
    for( i=0; i<m; ++i ) {
        scanf( "%d %d %d", &a, &b, &c );
        if( c > mp[a][b] ) continue;
        if( mp[a][b] == 1000000 ) {
            ns[cnt].a = a;
            ns[cnt++].b = b;
            vec[a].push_back(b);
            vec[b].push_back(a);
        }
        mp[a][b] = mp[b][a] = c;
    }
    /*
    for( i=1; i<=n; ++i ) {
        for( j=1; j<=n; ++j ) {
            printf( "%10d ", mp[i][j]);
        }
        printf( "\n" );
    }*/
}


int spfa( int from, int to ) {
    int Min[102], i, j, u, v, sz;
    bool mark[102];
    queue<int> Q;
    for( i=0; i<=n; ++i ) {
        Min[i] = 1000000;
        mark[i] = 0;
    }
    Min[from] = 0;
    Q.push( from );
    mark[from] = 1;
    while( !Q.empty() ) {
        u = Q.front();
        Q.pop();
        mark[u] = 0;
        sz = vec[u].size();
        for( i=0; i<sz; ++i ) {
            v = vec[u][i];
            if( mp[u][v] == 1000000 ) continue;
            if( Min[v] > Min[u] + mp[u][v] ) {
                Min[v] = Min[u] + mp[u][v];
                if( !mark[v] ) {
                    Q.push(v);
                    mark[v] = 1;
                }
            }
        }
    }

    return Min[to];

}



void solve() {
    int i, j, from, to, tp, temp;
    ans = 1000000;
    for( i=0; i<cnt; ++i ) {
        from = ns[i].a;
        to = ns[i].b;
        tp = mp[from][to];
        mp[from][to] = mp[to][from] = 1000000;
        temp = spfa( from, to );
        if( temp + tp < ans ) ans = temp + tp;
        mp[from][to] = mp[to][from] = tp;
    }

    if( ans == 1000000 ) puts( "It's impossible." );
    else printf( "%d\n", ans );
}
        


int main() {
//    freopen( "c:/aaa.txt", "r", stdin );
    while( scanf("%d %d", &n, &m) != EOF ) {
        input();
        solve();
    }
    return 0;
}