[BZOJ1779][Usaco2010 Hol]Cowwar 奶牛战争（最大流）

题目描述

传送门

题解

一眼网络流啊。。。

对于所有的J，s->i,1

对于所有的T，i->t,1

将J和E再拆两个点xi,yi，连边xi->yi,1

对于每一个J能移动到的位置（J或E），连边i->xj,1

对于每一个位置（J或E）能攻击到的T，连边yi->j,1

跑最大流即可

即用流来模拟了攻击的过程

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
#define N 160000
#define inf 2100000000

int n,m,s,t,maxflow;bool e[1005][1005];
char str
;
int tot,point
,nxt
,v
,remain
;
int deep
,last
,cur
,num
;
queue <int> q;

void addedge(int x,int y,int cap)
{
++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap;
++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0;
}
void bfs(int t)
{
for (int i=1;i<=t;++i) deep[i]=t;
deep[t]=0;
for (int i=1;i<=t;++i) cur[i]=point[i];
q.push(t);
while (!q.empty())
{
int now=q.front();q.pop();
for (int i=point[now];i!=-1;i=nxt[i])
if (deep[v[i]]==t&&remain[i^1])
{
deep[v[i]]=deep[now]+1;
q.push(v[i]);
}
}
}
int addflow(int s,int t)
{
int now=t,ans=inf;
while (now!=s)
{
ans=min(ans,remain[last[now]]);
now=v[last[now]^1];
}
now=t;
while (now!=s)
{
remain[last[now]]-=ans;
remain[last[now]^1]+=ans;
now=v[last[now]^1];
}
return ans;
}
void isap(int s,int t)
{
bfs(t);
for (int i=1;i<=t;++i) ++num[deep[i]];

int now=s;
while (deep[s]<t)
{
if (now==t)
{
maxflow+=addflow(s,t);
now=s;
}

bool has_find=0;
for (int i=cur[now];i!=-1;i=nxt[i])
if (deep[v[i]]+1==deep[now]&&remain[i])
{
has_find=1;
cur[now]=i;
last[v[i]]=i;
now=v[i];
break;
}

if (!has_find)
{
int minn=t-1;
for (int i=point[now];i!=-1;i=nxt[i])
if (remain[i]) minn=min(minn,deep[v[i]]);
if (!(--num[deep[now]])) break;
++num[deep[now]=minn+1];
cur[now]=point[now];
if (now!=s) now=v[last[now]^1];
}
}
}
int main()
{
scanf("%d%d",&n,&m);
scanf("%s",str+1);
for (int i=1;i<=m;++i)
{
int x,y;scanf("%d%d",&x,&y);
e[x][y]=e[y][x]=1;
}
s=3*n+1,t=s+1;
tot=-1;memset(point,-1,sizeof(point));
for (int i=1;i<=n;++i)
{
if (str[i]=='J')
{
addedge(s,i,1);
addedge(i,n+i,1);
addedge(n+i,n+n+i,1);
for (int j=1;j<=n;++j)
if (e[i][j]&&str[j]!='T') addedge(i,n+j,1);
}
if (str[i]=='T')
{
addedge(i,t,1);
for (int j=1;j<=n;++j)
if (e[i][j]&&str[j]!='T') addedge(n+n+j,i,1);
}
if (str[i]=='E') addedge(n+i,n+n+i,1);
}
isap(s,t);
printf("%d\n",maxflow);
}