解题报告 之 SGU326 Perspective

Description

Breaking news! A Russian billionaire has bought a yet undisclosed NBA team. He‘s planning to invest huge effort and money into making that team the best. And in fact he‘s been very specific about the expected result: the first place.

Being his advisor, you need to determine whether it‘s possible for your team to finish first in its division or not.

More formally, the NBA regular season is organized as follows: all teams play some games, in each game one team wins and one team loses. Teams are grouped into divisions, some games are between the teams in the same division, and some are between the teams
in different divisions.

Given the current score and the total number of remaining games for each team of your division, and the number of remaining games between each pair of teams in your division, determine if it‘s possible for your team to score at least as much wins as any other
team in your division.

Input

The first line of input contains N (2 ≤ N ≤ 20) — the number of teams in your division. They are numbered from 1 to N, your team has number 1.

The second line of input contains N integers w₁, w₂,..., w_N, where w_i is the total number of games that i^th team has won
to the moment.

The third line of input contains N integers r₁, r₂,..., r_N, where r_i is the total number of remaining games for the i^th team
(including the games inside the division).

The next N lines contain N integers each. The j^th integer in the i^th line of those contains a_ij — the number of games remaining between teams i and j.
It is always true that a_ij=a _ji and a_ii=0, for all ia_i1 + a_i2 +... + a_iN ≤ r_i.

All the numbers in input are non-negative and don‘t exceed 10\,000.

Output

On the only line of output, print "

YES

" (without quotes) if it‘s possible for the team 1 to score at least as much wins as any other team of its division, and "

NO

" (without quotes) otherwise.

Sample Input

sample input

sample output

3
1 2 2
1 1 1
0 0 0
0 0 0
0 0 0

YES

sample input

sample output

3
1 2 2
1 1 1
0 0 0
0 0 1
0 1 0

NO

题目大意：本小组内有N支队伍，你是team1。现在每个队已经有了score[i]的分，并且总共还剩下remain[i]场比赛（包括组内比赛和跨组比赛）。然后给你一个N*N的矩阵match表示组内剩余比赛，match[i][j]表示i与j队伍还剩多少场比赛，保证match[i][j]==match[j][i]，并且match[i][i]==0（忽然想起了前面那道坑题不保证这一点，妈蛋）。每场比赛不存在平局，赢的得1分，输的得2分。然后问你你们组是否有可能得小组第一（分数大于等于其他所有组内队伍）？

分析：更前面的WHU1124 Football Coach异曲同工。思路就不赘述了，请参考上一篇博文，细节在于这道题有外部比赛，那么我们的最优假设基于team1所有外部比赛都获胜，而其他组的所有外部比赛都输。而且组内比赛只要关于team1的也是都由team1获胜。剩下的再跑最大流限定得分为score[1]-score[i]，再验证满流即可。

上代码：

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

const int MAXN = 610;
const int MAXM = 361000;
const int INF = 0x3f3f3f3f;

struct Edge
{
	int to, cap, next;
};

Edge edge[MAXM];
int level[MAXN];
int head[MAXN];
int score[MAXN];
int remain[MAXN];
int match[MAXN][MAXN];
int src, des, cnt;

void addedge( int from, int to, int cap )
{
	edge[cnt].to = to;
	edge[cnt].cap = cap;
	edge[cnt].next = head[from];
	head[from] = cnt++;

	swap( from, to );

	edge[cnt].to = to;
	edge[cnt].cap = 0;
	edge[cnt].next = head[from];
	head[from] = cnt++;

}

int bfs()
{
	memset( level, -1, sizeof level );
	queue<int> q;
	while (!q.empty())
		q.pop();

	level[src] = 0;
	q.push( src );

	while (!q.empty())
	{
		int u = q.front();
		q.pop();

		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].to;
			if (edge[i].cap > 0 && level[v] == -1)
			{
				level[v] = level[u] + 1;
				q.push( v );
			}
		}
	}
	return level[des] != -1;
}

int dfs( int u, int f )
{
	if (u == des) return f;
	int tem;

	for (int i = head[u]; i != -1; i=edge[i].next)
	{
		int v = edge[i].to;
		if (edge[i].cap > 0 && level[v] == level[u] + 1)
		{
			tem = dfs( v, min( f, edge[i].cap ) );
			if (tem > 0)
			{
				edge[i].cap -= tem;
				edge[i^1].cap += tem;
				return tem;
			}
		}
	}
	level[u] = -1;
	return 0;
}

int Dinic()
{
	int ans = 0, tem;
	while (bfs())
	{
		while (tem = dfs( src, INF ))
		{
			ans += tem;
		}
	}
	return ans;
}

int main()
{
	int n;
	src = 0;
	des = 605;
	while (cin >> n)
	{
		memset( head, -1, sizeof head );
		cnt = 0;
		int flow = 0;
		for (int i = 1; i <= n; i++)
		{
			cin >> score[i];
		}
		for (int i = 1; i <= n; i++)
		{
			cin >> remain[i];
		}
		//每队还剩多少场比赛

		for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
		{
			cin >> match[i][j];
			remain[i] -= match[i][j];
		}
		//算出每个队伍有几场外部比赛

		int g = 21;
		for (int i = 1; i <= n; i++)
		{
			for (int j = i + 1; j <= n; j++)
			{
				if (match[i][j])
				{
					if (i == 1)
					{
						score[1] += match[i][j];
					}
					else
					{
						addedge( src, g, match[i][j] );
						addedge( g, i, match[i][j] );
						addedge( g, j, match[i][j] );
						g++;
						flow += match[i][j];
					}
				}
			}
		}

		int flag = true;
		for (int i = 1; i <= n; i++)
		{
			if(i==1)
				score[i] += remain[i];

			if (score[i] > score[1] && i > 1)
			{
				flag = false;
				break;
			}
		}//更新外部比赛后的最高得分
		if (!flag)
		{
			cout << "NO" << endl;
			continue;
		}

		for (int i = 2; i <= n; i++)
		{
			addedge( i, des, score[1] - score[i] );
		}

		if (Dinic() == flow) cout << "YES" << endl;
		else cout << "NO" << endl;

	}	

	return 0;
}

好了下午要复习一下嵌入式系统。。今天没听课，呵呵。