uva 1627

来源：互联网发布：考研数学网络课程编辑：程序博客网时间：2024/05/22 14:12

刘汝佳UVA1627代码，有一些困惑，看了很久才看懂，附带自己的理解在代码中。

// UVa1627 Team them up!

// Rujia Liu
#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

const int maxn = 100 + 5;

int n, G[maxn][maxn], color[maxn], diff[maxn], cc;
vector<int> team[maxn][2]; // team[cc][c] is the list of people in connected-component cc, color c

// returns false if not bipartite graph
bool dfs(int u, int c) {
color[u] = c;
team[cc][c-1].push_back(u);
for(int v = 0; v < n; v++) {//实际只能通过不满足联接的条件，来确定另外一个队的点集
    if(u != v && !(G[u][v] && G[v][u])) { // u and v must be in different groups
      if(color[v] > 0 && color[v] == color[u]) return false;
      if(!color[v] && !dfs(v, 3-c)) return false;
    }
}
return true;
}

bool build_graph() {
memset(color, 0, sizeof(color));
cc = 0; // current connected-component
for(int i = 0; i < n; i++)
    if(!color[i]) {
      team[cc][0].clear();

team[cc][1].clear();

if(!dfs(i, 1)) return false; //可以保证color[i]为0时，设i为1队并不与已有条件冲突

diff[cc] = team[cc][0].size() - team[cc][1].size();

      cc++;
    }

return true;
}

// d[i][j+n] = 1 iff we can arrange first i cc so that team 1 has j more people than team 2.
int d[maxn][maxn*2], teamno[maxn];

void print(int ans) {
vector<int> team1, team2;
for(int i = cc-1; i >= 0; i--) {
    int t;
    if(d[i][ans-diff[i]+n]) { t = 0; ans -= diff[i]; }
    else { t = 1; ans += diff[i]; }
    for(int j = 0; j < team[i][t].size(); j++)
      team1.push_back(team[i][t][j]);
    for(int j = 0; j < team[i][1^t].size(); j++)
      team2.push_back(team[i][1^t][j]);
}
printf("%d", team1.size());
for(int i = 0; i < team1.size(); i++) printf(" %d", team1[i]+1);
printf("\n");

printf("%d", team2.size());
for(int i = 0; i < team2.size(); i++) printf(" %d", team2[i]+1);
printf("\n");
}

void dp() {
memset(d, 0, sizeof(d));
d[0][0+n] = 1;
for(int i = 0; i < cc; i++)
    for(int j = -n; j <= n; j++) if(d[i][j+n]) {
      d[i+1][j+diff[i]+n] = 1;
      d[i+1][j-diff[i]+n] = 1;
    }

//每一次安排（即cc++），都是通过目前与所有的其他点的安排没有冲突的点来生成的，所\
以不论+diff[i]或-diff[i]都没问题

for(int ans = 0; ans <= n; ans++) {

    if(d[cc][ans+n]) { print(ans); return; }
    if(d[cc][-ans+n]) { print(-ans); return; }
}
}

int main() {
int T;
cin >> T;
while(T--) {
    cin >> n;
    memset(G, 0, sizeof(G));
    for(int u = 0; u < n; u++) {
      int v;
      while(cin >> v && v) G[u][v-1] = 1;
    }

    if(n == 1 || !build_graph()) cout << "No solution\n";
    else dp();

    if(T) cout << "\n";
}
return 0;
}

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