UVA1391/LA3713 Astronauts

题意：有A、B、C3个任务分配给n个宇航员，其中每个宇航员恰好分配一个任务。假设n个宇航员的平均年龄为x，只有年龄大于x的才能领取A任务；只有年龄严格小于x的才能领取B任务，而任务C没有限制。有m对宇航员相互讨厌，因此不能分配同一任务。求出是否能找出符合的任务方案。（转自http://blog.csdn.net/u011345461/article/details/39779721）

题目看上去是ABC三个选择，实际上每个人只有两个选择。对于每对矛盾，如果两个人选择相同（即均为BC或均为AC），那么一个选C另一个必选B（A），另一个选C一个必选B（A），一个选B（A）另一个必选C，另一个选B（A）一个必选C；如果两个人选择不同（一个AC一个BC），那么一个选C另一个必选B，另一个选C一个必选A。

这题数据范围很大，好奇nm做法是怎么水过去的。。

正解是tarjan缩点，标记每个块的对立块，建立反图，对反图top排序，对于当前的块x,删除对立块和对立块在反图上的出边，并把当前快入队。最后看队列里没被删除的块即为答案块。把点扫一扫看看在不在答案块，在的话这个点的值代表选A（B）还是选C就是这个人的选择

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <queue>
 #include <vector>
 #include <map>
 #include <string>
 #include <cmath>
 #define min(a, b) ((a) < (b) ? (a) : (b))
 #define max(a, b) ((a) > (b) ? (a) : (b))
 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
 template<class T>
 inline void swap(T &a, T &b)
 {
     T tmp = a;a = b;b = tmp;
 }
 inline void read(int &x)
 {
     x = ;char ch = getchar(), c = ch;
     while(ch < '' || ch > '') c = ch, ch = getchar();
     while(ch <= '' && ch >= '') x = x *  + ch - '', ch = getchar();
     if(c == '-') x = -x;
 }
 const int INF = 0x3f3f3f3f;
 const int MAXN =  + ;
 struct Edge
 {
     int u,v,nxt;
     Edge(int _u, int _v, int _nxt){u = _u;v = _v;nxt = _nxt;}
     Edge(){}
 }edge1[MAXN << ], edge2[MAXN << ];
 int head1[MAXN], head2[MAXN], vis[MAXN], tag[MAXN], rebelong[MAXN], del[MAXN], q[MAXN], he, ta, indeg[MAXN], cnt1, cnt2, n, m, belong[MAXN], group, b[MAXN], bb[MAXN], dfn[MAXN], stack[MAXN], top, low[MAXN], dfst, year[MAXN], x, tmp1, tmp2;
 inline void insert1(int a, int b){edge1[++ cnt1] = Edge(a, b, head1[a]), head1[a] = cnt1;}
 inline void insert2(int a, int b){edge2[++ cnt2] = Edge(a, b, head2[a]), head2[a] = cnt2;}
 void dfs(int u)
 {
     b[u] = bb[u] = , dfn[u] = low[u] = ++ dfst, stack[++ top] = u;
     for(int pos = head1[u];pos;pos = edge1[pos].nxt)
     {
         int v = edge1[pos].v;
         if(!b[v]) dfs(v), low[u] = min(low[u], low[v]);
         else if(bb[v]) low[u] = min(low[u], dfn[v]);
     }
     if(low[u] == dfn[u])
     {
         int now = -;++ group;
         while(now != u) now = stack[top --], bb[now] = , belong[now] = group;
     }
 }
 void rebuild()
 {
     memset(indeg, , sizeof(indeg)), memset(head2, , sizeof(head2));
     for(int i = ;i <= cnt1;++ i)
         if(belong[edge1[i].u] != belong[edge1[i].v])
             insert2(belong[edge1[i].v], belong[edge1[i].u]), ++ indeg[belong[edge1[i].u]];
 }
 void tarjan()
 {
     memset(belong, , sizeof(belong)), memset(dfn, , sizeof(dfn)), memset(low, , sizeof(low)), group = , memset(b, , sizeof(b)), memset(bb, , sizeof(bb)), dfst = , top = , memset(rebelong, , sizeof(rebelong));
     for(int i = ;i <= n;++ i) if(!b[i << ]) dfs(i << );
     for(int i = ;i <= n;++ i) if(!b[i <<  | ]) dfs(i <<  | );
     for(int i = ;i <= n;++ i) rebelong[belong[i << ]] = belong[i <<  | ], rebelong[belong[i <<  | ]] = belong[i << ];
     rebuild();
 }
 void top_sort()
 {
     he = , ta = , memset(del, , sizeof(del));
     for(int i = ;i <= group;++ i) if(!indeg[i]) q[ta ++] = i;
     while(he < ta)
     {
         int now = q[he ++], renow = rebelong[now];
         if(del[now]) continue; del[renow] = ;
         for(int pos = head2[renow];pos;pos = edge2[pos].nxt)
         {
             int v = edge2[pos].v;
             del[v] = ;
             for(int poss = head2[v];poss;poss = edge2[poss].nxt) -- indeg[edge2[poss].v];
         }
         for(int pos = head2[now];pos;pos = edge2[pos].nxt)
         {
             int v = edge2[pos].v;
             if(del[v]) continue;
             -- indeg[v];
             if(!indeg[v]) q[ta ++] = v;
         }
     }
 }
 int main()
 {
     while(scanf("%d %d", &n, &m) != EOF && n && m)
     {
         cnt1 = , memset(head1, , sizeof(head1)), x = ;
         for(int i = ;i <= n;++ i) read(year[i]), x += year[i];
         if(x % n == ) x = x / n;
         else x = x / n + ;
         for(int i = ;i <= n;++ i) year[i] = year[i] >= x;  //year[i] = 1 表示选A或C  year[i] = 0 表示选B或C
         for(int i = ;i <= m;++ i)
         {
             read(tmp1), read(tmp2);
             if(year[tmp1] == year[tmp2]) insert1(tmp1 <<  | , tmp2 << ), insert1(tmp2 <<  | , tmp1 << ), insert1(tmp1 << , tmp2 <<  | ), insert1(tmp2 << , tmp1 <<  | );
             else insert1(tmp1 <<  | , tmp2 << ), insert1(tmp2 <<  | , tmp1 << );
         }
         tarjan();
         int flag = ;
         for(int i = ;i <= n;++ i) if(belong[i <<  | ] == belong[i << ]) flag = ;
         if(flag)
         {
             printf("No solution.\n");
             continue;
         }
         top_sort();
         memset(vis, , sizeof(vis)), memset(tag, , sizeof(tag));
         for(int i = ;i < ta;++ i) if(!del[q[i]]) vis[q[i]] = ;
         for(int i = ;i <= n;++ i)
             if(vis[belong[i << ]]) tag[i] = ;
             else tag[i] = ;
         for(int i = ;i <= n;++ i)
             if(year[i] && !tag[i]) printf("A\n");
             else if(!year[i] && !tag[i]) printf("B\n");
             else printf("C\n");
     }
     return ;
 }

UVA1391/LA3713