1198 - Karate Competition
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PDF (English) | Statistics | Forum |
| Time Limit: 2 second(s) | Memory Limit: 32 MB |
Your karate club challenged another karate club in your town. Each club enters N players into the match, and each player plays one game against a player from the other team. Each game that is won is worth 2 points, and each game that is drawn is worth 1 point. Your goal is to score as many points as possible.
Your secret agents have determined the skill of every member of the opposing team, and of course you know the skill of every member of your own team. You can use this information to decide which opposing player will play against each of your players in order to maximize your score. Assume that the player with the higher skill in a game will always win, and if the players have the same skill then they will draw.
You will be given the skills of your players and of the opposing players, you have to find the maximum number of points that your team can score.
Input
Input starts with an integer T (≤ 70), denoting the number of test cases.
Each case starts with a line containing an integer N (1 ≤ N ≤ 50). The next line contains N space separated integers denoting the skills of the players of your team. The next line also contains N space separated integers denoting the skills of the players of the opposite team. Each of the skills lies in the range [1, 1000].
Output
For each case, print the case number and the maximum number of points your team can score.
Sample Input |
Output for Sample Input |
| 4 2 4 7 6 2 2 6 2 4 7 3 5 10 1 5 10 1 4 10 7 1 4 15 3 8 7 |
Case 1: 4 Case 2: 2 Case 3: 4 Case 4: 5 |
题目大意：你的空手道馆要跟别的空手道馆打比赛，每个人都有个技能点，要是技能点比对手的技能点高就加2分，平手加1分，输了不加分。问最多能得多少分
感觉能贪心做，类似于田忌赛马的思路，但是竟然放在了费用流就用费用流做吧..n^2建图，跑费用流即可
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=1010;
const int maxm=10010;
const int inf=0x3f3f3f3f;
struct Node
{
int to;
int capa;
int cost;
int next;
}edge[maxm];
int cnt;
int source,sink;
int head[maxn];
int numa[maxn];
int numb[maxn];
int dis[maxn];
bool vis[maxn];
int rec[maxn];
int pre[maxn];
int ldj;
void init()
{
memset(head,-1,sizeof(head));
cnt=0;
return;
}
void add(int u,int v,int capa,int cost)
{
edge[cnt].to=v;
edge[cnt].capa=capa;
edge[cnt].cost=cost;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].to=u;
edge[cnt].capa=0;
edge[cnt].cost=-cost;
edge[cnt].next=head[v];
head[v]=cnt++;
return;
}
bool spfa()
{
memset(vis,false,sizeof(vis));
memset(dis,inf,sizeof(dis));
memset(rec,-1,sizeof(rec));
memset(pre,-1,sizeof(pre));
queue<int> que;
que.push(source);
dis[source]=0;
vis[source]=true;
while(!que.empty())
{
int node=que.front();
que.pop();
vis[node]=false;
for(int i=head[node];~i;i=edge[i].next)
{
int v=edge[i].to;
if(edge[i].capa>0&&dis[v]>dis[node]+edge[i].cost)
{
dis[v]=dis[node]+edge[i].cost;
rec[v]=i;
pre[v]=node;
if(!vis[v])
{
vis[v]=true;
que.push(v);
}
}
}
}
return dis[sink]!=inf;
}
int mcmf()
{
int maxflow=0;
int mincost=0;
while(spfa())
{
int node=sink;
int flow=inf;
while(node!=source)
{
flow=min(flow,edge[rec[node]].capa);
node=pre[node];
}
node=sink;
while(node!=source)
{
mincost+=flow*edge[rec[node]].cost;
edge[rec[node]].capa-=flow;
edge[rec[node]^1].capa+=flow;
node=pre[node];
}
}
return -mincost;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int test;
scanf("%d",&test);
for(int cas=1;cas<=test;cas++)
{
init();
int n;
scanf("%d",&n);
source=0;
sink=n*2+1;
for(int i=1;i<=n;i++)
{
scanf("%d",&numa[i]);
}
for(int i=1;i<=n;i++)
{
scanf("%d",&numb[i]);
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(numa[i]<numb[j])
{
add(i,n+j,1,0);
}
else if(numa[i]==numb[j])
{
add(i,n+j,1,-1);
}
else
{
add(i,n+j,1,-2);
}
}
}
for(int i=1;i<=n;i++)
{
add(source,i,1,0);
add(n+i,sink,1,0);
}
printf("Case %d: %d\n",cas,mcmf());
}
return 0;
}