1001: [BeiJing2006]狼抓兔子
Time Limit: 15 Sec Memory Limit: 162 MBSubmit: 22666 Solved: 5704
[ Submit][ Status][ Discuss]
Description
Input
Output
输出一个整数,表示参与伏击的狼的最小数量.
Sample Input
5 6 4
4 3 1
7 5 3
5 6 7 8
8 7 6 5
5 5 5
6 6 6
Sample Output
HINT
2015.4.16新加数据一组,可能会卡掉从前可以过的程序。
Source
最小割模板题,难点在于将原图转化成对偶图
剩下的跑一发最短路即可
#include<set>
#include<queue>
#include<math.h>
#include<vector>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long ll;
#define maxn 2000005
struct node
{
int c,v;
};
queue<node>t;
vector<node>q[maxn];
int best[maxn],flag[maxn];
int main(void)
{
int n,m,i,j,x,ans;
while(scanf("%d%d",&n,&m)!=EOF)
{
node now;
memset(flag,0,sizeof(flag));
if(n==1 || m==1)
{
ans=2147483647;
for(i=1;i<=max(n-1,m-1);i++)
{
scanf("%d",&x);
ans=min(ans,x);
}
printf("%d\n",ans);
continue;
}
for(i=0;i<=(n-1)*(m-1)*2+1;i++)
q[i].clear();
for(i=1;i<=n;i++)
for(j=1;j<=m-1;j++)
{
scanf("%d",&x);
now.c=x;
if(i==1)
{
now.v=0;
q[(i-1)*(m-1)*2+j].push_back(now);
now.v=(i-1)*(m-1)*2+j;
q[0].push_back(now);
}
else if(i==n)
{
now.v=(n-1)*(m-1)*2+1;
q[(2*i-3)*(m-1)+j].push_back(now);
now.v=(2*i-3)*(m-1)+j;
q[(n-1)*(m-1)*2+1].push_back(now);
}
else
{
now.v=(i-1)*(m-1)*2+j;
q[(2*i-3)*(m-1)+j].push_back(now);
now.v=(2*i-3)*(m-1)+j;
q[(i-1)*(m-1)*2+j].push_back(now);
}
}
for(i=1;i<=n-1;i++)
for(j=1;j<=m;j++)
{
scanf("%d",&x);
now.c=x;
if(j==1)
{
now.v=(n-1)*(m-1)*2+1;
q[(i-1)*2*(m-1)+m].push_back(now);
now.v=(i-1)*2*(m-1)+m;
q[(n-1)*(m-1)*2+1].push_back(now);
}
else if(j==m)
{
now.v=0;
q[(2*i-1)*(m-1)].push_back(now);
now.v=(2*i-1)*(m-1);
q[0].push_back(now);
}
else
{
now.v=(2*i-1)*(m-1)+j-m;
q[(2*i-1)*(m-1)+j].push_back(now);
now.v=(2*i-1)*(m-1)+j;
q[(2*i-1)*(m-1)+j-m].push_back(now);
}
}
for(i=1;i<=n-1;i++)
for(j=1;j<=m-1;j++)
{
scanf("%d",&x);
now.c=x;
now.v=2*(i-1)*(m-1)+j;
q[(2*i-1)*(m-1)+j].push_back(now);
now.v=(2*i-1)*(m-1)+j;
q[2*(i-1)*(m-1)+j].push_back(now);
}
for(i=0;i<=(n-1)*(m-1)*2+1;i++)
best[i]=1000000000;
flag[0]=1;best[0]=0;
now.c=0;now.v=0;
t.push(now);
while(t.empty()==0)
{
now=t.front();
t.pop();
flag[now.v]=0;
for(i=0;i<q[now.v].size();i++)
{
node tmp=q[now.v][i];
if(best[tmp.v]>best[now.v]+tmp.c)
{
best[tmp.v]=best[now.v]+tmp.c;
if(flag[tmp.v]==0)
{
flag[tmp.v]=1;
t.push(tmp);
}
}
}
}
printf("%d\n",best[(n-1)*(m-1)*2+1]);
}
return 0;
}