ZOJ3508_The War(网络流最大流)

解题报告

http://blog.csdn.net/juncoder/article/details/38235609

题目传送门

题意：

N个士兵，M个武器，每个士兵能接受的武器重量范围是[minw，maxw]

思路：

本来以为二分图可以的，（看错数据范围了，，，）贪心好像可以。

scf说网络流可以缩点。

建图方式：源点和士兵连一条线，每个士兵与[1,1000]的武器重量连边，[1,1000]与汇点连线，容量是武器i的数量

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define inf 99999999
int w,n,m,head[4010],l[4010],vis[4010],d[4010],cnt;
struct node1
{
    int maxx,minn;
}we[4000];
struct node
{
    int v,w,next;
}edge[2510000];
void add(int u,int v,int w)
{
    edge[cnt].v=v;
    edge[cnt].w=w;
    edge[cnt].next=head[u];
    head[u]=cnt++;
    edge[cnt].v=u;
    edge[cnt].w=0;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}
int bfs()
{
    queue<int>Q;
    Q.push(0);
    memset(l,-1,sizeof(l));
    l[0]=0;
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v;
            if(l[v]==-1&&edge[i].w)
            {
                l[v]=l[u]+1;
                Q.push(v);
            }
        }
    }
    if(l[n+1000+1]>0)
    return 1;
    else return 0;
}
int dfs(int x,int f)
{
    int i,a;
    if(x==n+1000+1)return f;
    for(i=head[x];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(edge[i].w&&l[v]==l[x]+1&&(a=dfs(v,min(f,edge[i].w))))
        {
            edge[i].w-=a;
            edge[i^1].w+=a;
            return a;
        }
    }
    l[x]=-1;
    return 0;
}
int main()
{
    int i,j,s,e;
    while(~scanf("%d%d",&n,&m))
    {
        cnt=0;
        memset(d,0,sizeof(d));
        memset(edge,0,sizeof(edge));
        memset(head,-1,sizeof(head));
        for(i=1;i<=n;i++)
        {
            add(0,i,1);
            scanf("%d%d",&we[i].minn,&we[i].maxx);
        }
        for(i=1;i<=m;i++)
        {
            scanf("%d",&w);
            d[w]++;
        }
        for(i=1;i<=1000;i++)
        {
            add(i+n,n+1000+1,d[i]);
            for(j=1;j<=n;j++)
            {
                if(i<=we[j].maxx&&i>=we[j].minn)
                {
                    add(j,i+n,1);
                }
            }
        }
        int ans=0,a;
        while(bfs())
            while(a=dfs(0,inf))
            ans+=a;
        printf("%d\n",ans);
    }
}

The War

Input

There multiple test cases. The first line of each case are two integers N, M. Then the following N lines, each line contain two integers minWi, maxWi for each soldier. Next M lines, each
line contain one integer W represents the weight of each weapon.

Output

For each case, output one integer represents the maximal number of armed soldier you can get.

Sample Input

3 3
1 5
3 7
5 10
4
8
9
2 2
5 10
10 20
4
21

Sample Output

2
0

ZOJ3508_The War(网络流最大流)