sgu 326——Perspective

A - Perspective

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Time Limit:500MS    
Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Status
Practice
SGU 326

 

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,...,     wN, where
wi is the total number of games that
ith team has won to the moment.    

The third line of input contains N integers r1,    
r2,...,     rN, where
ri is the total number of remaining games for the
ith team (including the games inside the division).    

The next N lines contain N integers each. The jth integer in the
ith line of those contains
aij — the number of games remaining between teams
i and j. It is always true that aij=a
ji and aii=0, for all
iai1 +     ai2 +... +    
aiN ≤     ri.    

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
31 2 21 1 10 0 00 0 00 0 0	YES

sample input	sample output
31 2 21 1 10 0 00 0 10 1 0	NO

 

網路流

 

#include <iostream>#include<cstdio>#include<cstring>#include<cstdio>#include<queue>#include<vector>using namespace std;#define INF 100000001const int maxint=1000001;int n,m,s,t;int num[100],remain[100];int map[100][100];struct edgee{int from,to,cap,flow;};vector<edgee> edges;vector<int> G[10003];bool vis[10003];int dist[10003];int cur[10003];void add(int from,int to,int cap){edges.push_back((edgee){from,to,cap,0});edges.push_back((edgee){to,from,0,0});int k=edges.size();G[from].push_back(k-2);//奇數為正向弧 G[to].push_back(k-1);}int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a<b?a:b;}bool bfs(){memset(vis,0,sizeof(vis));queue<int> q;dist[s]=0;vis[s]=1;q.push(s);while(!q.empty()){int u=q.front();q.pop();int sz=G[u].size();for(int i=0;i<sz;i++){edgee& e=edges[G[u][i]];if(!vis[e.to] && e.cap>e.flow){vis[e.to]=1;dist[e.to]=dist[u]+1;q.push(e.to);}}}return vis[t];}int dfs(int u,int low){if(u==t || low==0)return low;int flow=0,sz=G[u].size(),d;//&引用的妙用for(int& i=cur[u];i<sz;i++) //正在考慮的最後一條邊開始考慮（效率關鍵）{edgee& e=edges[G[u][i]];if(dist[u]+1==dist[e.to] && (d=dfs(e.to,min(e.cap-e.flow,low)))>0 ){e.flow+=d;edges[G[u][i]^1].flow-=d;flow+=d;low-=d;if(low==0)break;}}return flow;}int Dinic(){int flow=0;while(bfs()){memset(cur,0,sizeof(cur));flow+=dfs(s,maxint);}return flow;}int main(){int i,j;s=4003,t=4004;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&num[i]);for(int i=1;i<=n;i++)scanf("%d",&remain[i]);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)scanf("%d",&map[i][j]);num[1]+=remain[1];for(int i=1;i<=n;i++)if(map[1][i]!=0)map[1][i]=map[i][1]=0;for(int i=2;i<=n;i++){add(i,t,INF);if(num[1]-num[i]<0){printf("NO\n");return 0;}add(i+100,i,num[1]-num[i]);}int k=0;int sum=0;for(int i=2;i<=n;i++)for(int j=2;j<=n;j++)if(map[i][j]!=0){k++;sum+=map[i][j];add(300+k,i+100,INF);add(300+k,j+100,INF);add(s,300+k,map[i][j]);map[i][j]=map[j][i]=0;}int temp=Dinic();if(temp==sum)   printf("YES\n");elseprintf("NO\n");return 0;}