| The Maximum Number of Strong Kings
Description A tournament can be represented by a complete graph in which each vertex denotes a player and a directed edge is from vertex x to vertex y if player x beats player y. For a player x in a tournament T, the score of x is the number of players beaten by x. The score sequence of T, denoted by S(T) = (s1, s2, . . . , sn), is a non-decreasing list of the scores of all the players in T. It can be proved that S(T) = (s1, s2, . . . , sn) is a score sequence of T if and only if  for k = 1, 2, . . . , n and equality holds when k = n. A player x in a tournament is a strong king if and only if x beats all of the players whose scores are greater than the score of x. For a score sequence S, we say that a tournament T realizes S if S(T) = S. In particular, T is a heavy tournament realizing S if T has the maximum number of strong kings among all tournaments realizing S. For example, see T2 in Figure 1. Player a is a strong king since the score of player a is the largest score in the tournament. Player b is also a strong king since player b beats player a who is the only player having a score larger than player b. However, players c, d and e are not strong kings since they do not beat all of the players having larger scores. The purpose of this problem is to find the maximum number of strong kings in a heavy tournament after a score sequence is given. For example,Figure 1 depicts two possible tournaments on five players with the same score sequence (1, 2, 2, 2, 3). We can see that there are at most two strong kings in any tournament with the score sequence (1, 2, 2, 2, 3) since the player with score 3 can be beaten by only one other player. We can also see that T2 contains two strong kings a and b. Thus, T2 is one of heavy tournaments. However, T1 is not a heavy tournament since there is only one strong king in T1. Therefore, the answer of this example is 2.  Input The first line of the input file contains an integer m, m <= 10, which represents the number of test cases. The following m lines contain m score sequences in which each line contains a score sequence. Note that each score sequence contains at most ten scores. Output The maximum number of strong kings for each test case line by line. Sample Input 5 1 2 2 2 3 1 1 3 4 4 4 4 3 3 4 4 4 4 5 6 6 6 0 3 4 4 4 5 5 5 6 0 3 3 3 3 3 Sample Output 2 4 5 3 5 |
建图：（s， i）为源点到每个人的流量，再把每场比赛看成一个点，到达这点的边可能有（i， v）， （j， v）两条流量为1的边，（v， t ）每场比赛和终点相连接，可以通过控制到v的边来限制只能那个点赢得比赛，因而能得出最大的kings数量。
#include <iostream>#include <sstream>#include <string>#include <cstdio>#include <cstring>#define maxn 100#define INF 2000000000#define max(a, b) a > b ? a : b#define min(a, b) a < b ? a : b using namespace std; int score[maxn];int front[maxn];int pre[maxn];int dis[maxn];int gap[maxn];int cur[maxn];int index;int n, total; struct Edge{ int v; int w; int next;}edge[1000]; void init(){ memset( pre, -1, sizeof(pre) ); index = 0;} void Add( int u, int v, int w ){ edge[index].v = v; edge[index].w = w; edge[index].next = pre[u]; pre[u] = index++; edge[index].v = u; edge[index].w = 0; edge[index].next = pre[v]; pre[v] = index++;} int maxflow_Sap( int s, int t ){ memset( dis, 0, sizeof(dis) ); memset( gap, 0, sizeof(gap) ); memcpy( cur, pre, sizeof(cur) ); int flow = INF, u = s, maxflow = 0, i; gap[0] = t+1; while ( dis[s] < t+1 ){ for ( i = cur[u]; i != -1; i = edge[i].next ){ if ( edge[i].w > 0 && dis[u] == dis[edge[i].v] + 1 ){ break; } } cur[u] = i; if ( cur[u] != -1 ){ flow = min( flow, edge[i].w ); front[edge[i].v] = u; u = edge[i].v; if ( u == t ){ maxflow += flow; for ( i = front[t]; ; i = front[i] ){ edge[cur[i]].w -= flow; edge[cur[i]^1].w += flow; if ( i == s ){ break; } } u = s; flow = INF; } } else{ int mindis = t + 1; for ( i = pre[u]; i != -1; i = edge[i].next ){ if ( edge[i].w > 0 && mindis > dis[edge[i].v] + 1 ){ mindis = dis[edge[i].v] + 1; cur[u] = i; } } gap[dis[u]]--; if ( !gap[dis[u]] ){ break; } gap[mindis]++; dis[u] = mindis; if ( u != s ){ u = front[u]; } } } return maxflow;} int Build( int s, int t ){ int ans = n-1; while ( ans >= 0 ){ init(); int cnt = n; for ( int i = 1; i <= n; i++ ){ Add( s, i, score[i] ); for ( int j = i+1; j <= n; j++ ){ cnt++; if ( i > ans && score[j] > score[i] ){ //限制只能i赢 Add( i, cnt, 1 ); } else{ //i，j都可能赢 Add( i, cnt, 1 ); Add( j, cnt, 1 ); } Add( cnt, t, 1 ); } } if ( n*(n-1)/2 != maxflow_Sap( s, t ) ){ return n-ans-1; } ans--; } return n;} int main(){ int T; scanf("%d%*c", &T); while ( T-- ){ char c[100]; gets(c); stringstream ss(c); int i = 1; n = 0; while ( ss >> score[i] ){ total += score[i++]; } n = i - 1; printf("%d\n", Build( 0, n*(n-1)/2+n+1 )); } return 0;}