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 https://odzkskevi.qnssl.com/9698e76abf2d161889085123803f9464?v=1508663873

 

【题解】

把“两个人不相互认识”连边，形成若干连通分量

不难发现每个联通分量进行黑白染色后，必定选黑或选白

DP[i][j]表示前i个连通块，选j个黑块是否可行

从dp[n][mid]开始向左右检查即可

 

因为边数开小以及偶数n的处理不当RE/WA了很多发

要用大数据测/对拍/手出数据验证程序的每一部分

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector> 
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))

inline void swap(int &a, int &b)
{
    int tmp = a;a = b;b = tmp;
}

inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9')c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
    if(c == '-')x = -x;
}

const int INF = 0x3f3f3f3f;
const int MAXN = 100 + 10;

struct Edge
{
    int u,v,nxt;
    Edge(int _u, int _v, int _nxt){u = _u;v = _v;nxt = _nxt;}
    Edge(){}
}edge[MAXN * MAXN << 1];
int head[MAXN], cnt;
inline void insert(int a, int b)
{
    edge[++cnt] = Edge(a,b,head[a]);
    head[a] = cnt;
}

int t,n,tot,x[MAXN],y[MAXN],b[MAXN],g[MAXN][MAXN],tag,dp[MAXN][MAXN],zhuanyi1[MAXN][MAXN], zhuanyi2[MAXN][MAXN], zhuanyi3[MAXN][MAXN];
std::vector<int> xx[MAXN], yy[MAXN];

int dfs(int u, int flag)
{
    b[u] = flag;
    if(flag == 1) ++ x[tot], flag = 2, xx[tot].push_back(u);
    else flag = 1, ++ y[tot], yy[tot].push_back(u);
    for(register int pos = head[u];pos;pos = edge[pos].nxt)
    {
        int v = edge[pos].v;
        if(!b[v])
        {
            if(dfs(v, flag))
                return 1;
        }
        else if(b[v] != flag) 
            return 1;
    }
    return 0;
}

void dfs2(int p, int q)
{
    if(p == 0) return;
    b[p] = zhuanyi3[p][q];
    dfs2(zhuanyi1[p][q], zhuanyi2[p][q]);
}

int check(int num)
{
    if(!dp[tot][num]) return 0;
    printf("%d ", num);
    memset(b, 0, sizeof(b));
    dfs2(tot, num);
    for(register int i = 1;i <= tot;++i)
        if(b[i] == 1)
            for(register int j = 0;j < xx[i].size();++ j)
                printf("%d ", xx[i][j]);
        else
            for(register int j = 0;j < yy[i].size();++ j)
                printf("%d ", yy[i][j]);
    printf("\n%d ", n - num);
    for(register int i = 1;i <= tot;++i)
        if(b[i] == 1)
            for(register int j = 0;j < yy[i].size();++ j)
                printf("%d ", yy[i][j]);
        else
            for(register int j = 0;j < xx[i].size();++ j)
                printf("%d ", xx[i][j]);
    putchar('\n');
    return 1;
}

void init() 
{
    tot = 0;cnt = 0;tag = 0;
    memset(head, 0, sizeof(head));
    memset(g, 0, sizeof(g));
    memset(b, 0, sizeof(b));
    memset(x, 0, sizeof(x));
    memset(y, 0, sizeof(y));
    memset(dp, 0, sizeof(dp));
    memset(zhuanyi1, 0, sizeof(zhuanyi1));
    memset(zhuanyi2, 0, sizeof(zhuanyi2));
    memset(zhuanyi3, 0, sizeof(zhuanyi3));
    for(register int i = 1;i <= n;++ i) xx[i].clear(), yy[i].clear();
}

int main()
{
    read(t);
    for(;t;--t)
    {
        read(n);
        init();
        for(register int i = 1;i <= n;++ i)
        {
            int tmp;
            read(tmp);
            while(tmp)
            {
                g[i][tmp] = 1;
                read(tmp);
            }
        }
        for(register int i = 1;i <= n;++ i)
            for(register int j = i + 1;j <= n;++ j)
                if(!g[i][j] || !g[j][i])
                    insert(i, j), insert(j, i);
        for(register int i = 1;i <= n;++ i)
            if(!b[i])
            {
                ++ tot;
                if(dfs(i, 1))
                {
                    printf("No solution\n\n");
                    tag = 1;
                    break;
                }
            }
        if(tag)continue;
        dp[0][0] = 1;
        //dp[i][j]表示前i个连通分量，标号为1的有j个是否可行 
        for(register int i = 1;i <= tot;++ i)
            for(register int j = n;j >= 0;-- j)
            {
                if(j - x[i] >= 0 && dp[i - 1][j - x[i]])
                {
                    zhuanyi1[i][j] = i - 1;
                    zhuanyi2[i][j] = j - x[i];
                    zhuanyi3[i][j] = 1;
                    dp[i][j] = 1;
                }
                else if(j - y[i] >= 0 && dp[i - 1][j - y[i]])
                {
                    zhuanyi1[i][j] = i - 1;
                    zhuanyi2[i][j] = j - y[i];
                    zhuanyi3[i][j] = 2;
                    dp[i][j] = 1;
                }
            }
        int mid = n/2;
        if(n & 1)
        {
            mid += 1;
            for(register int i = 0;;++ i)
            {
                if(!(i + mid <= n && mid - i > 0)) 
                {
                    tag = 1;    
                    break;
                }
                if(check(i + mid)) 
                {
                    tag = 1;
                    break;
                }
                if(check(mid - i)) 
                {
                    tag = 1;
                    break;
                }
            }
        }
        else
        {
            for(register int i = 1;i <= mid;++i)
            {            
                if(check(mid - i + 1)) 
                {
                    tag = 1;
                    break;
                }
                if(check(mid + i)) 
                {
                    tag = 1;
                    break;
                }
            }
        }
        if(!tag)printf("No solution\n");
        printf("\n");
    }
    return 0;
}