【BZOJ1001】【Beijing2006】狼抓兔子（平面图转对偶图：最小割+最短路）

题目描述

题解

代码

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

const int max_n=1005;
const int max_m=1005;
const int max_N=max_n*max_m*2+5;
const int max_e=3*max_n*max_m+5;
const int max_queue=max_N*2+5;
const int inf=2100000000;

int n,m,N,now,nowlast;
int heng[max_n][max_m],zong[max_n][max_m],xie[max_n][max_m];
int next[max_e*2],point[max_N],c[max_e*2],v[max_e*2],tot;
int dis[max_N],queue[max_queue+5],head,tail;
bool b[max_N];

inline void add(int x,int y,int val){
++tot; next[tot]=point[x]; point[x]=tot; v[tot]=y; c[tot]=val;
++tot; next[tot]=point[y]; point[y]=tot; v[tot]=x; c[tot]=val;
}

int main(){
scanf("%d%d",&n,&m);
//总点数
N=(n-1)*(m-1)*2+2;

//特判
if (n==1&&m==1){
printf("0\n");
return 0;
}
if (n==1){
int Min=inf,x;
for (int i=1;i<m;++i)
scanf("%d",&x),Min=min(Min,x);
printf("%d\n",Min);
return 0;
}
if (m==1){
int Min=inf,x;
for (int i=1;i<n;++i)
scanf("%d",&x),Min=min(Min,x);
printf("%d\n",Min);
return 0;
}

//read
for (int i=1;i<=n;++i)
for (int j=1;j<m;++j)
scanf("%d",&heng[i][j]);
for (int i=1;i<n;++i)
for (int j=1;j<=m;++j)
scanf("%d",&zong[i][j]);
for (int i=1;i<n;++i)
for (int j=1;j<m;++j)
scanf("%d",&xie[i][j]);

//build up
for (int i=1;i<n;++i)
for (int j=1;j<m;++j){
now=(i-1)*(m-1)+j;
//斜向
add(now*2,now*2+1,xie[i][j]);
//纵向
if (j!=1) add(now*2-1,now*2,zong[i][j]);
//横向
if (i!=1){
nowlast=now-(m-1);
add(nowlast*2,now*2+1,heng[i][j]);
}
}
//S
for (int i=1;i<n;++i){
now=(i-1)*(m-1)+1;
add(1,now*2,zong[i][1]);
now=(i-1)*(m-1)+m-1;
add(now*2+1,N,zong[i][m]);
}
//T
for (int i=1;i<m;++i){
now=i;
add(now*2+1,N,heng[1][i]);
now=(n-2)*(m-1)+i;
add(1,now*2,heng
[i]);
}

//SPFA
memset(dis,0x7f,sizeof(dis));
memset(b,0,sizeof(b));
dis[1]=0;
head=0; tail=1;
queue[tail]=1;
b[1]=true;
while (head!=tail){
head++;
now=queue[head%max_queue];
b[now]=false;
for (int i=point[now];i;i=next[i])
if (dis[v[i]]>dis[now]+c[i]){
dis[v[i]]=dis[now]+c[i];
if (!b[v[i]]){
tail++;
queue[tail%max_queue]=v[i];
b[v[i]]=true;
}
}
}
printf("%d\n",dis
);
}