BZOJ1001 [Beijing2006] 狼抓兔子

题解：

求平面图的最小割。

转成对偶图求最短路。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn=2000010;
const int inf=1e9;
int N,M;
int s,t;
struct e {
    int from;
    int to;
    int next;
    int flow;
}edge[maxn*4];
int head[maxn];
int tol;
void addedge (int u,int v,int flow) {
    edge[tol].from=u;
    edge[tol].to=v;
    edge[tol].next=head[u];
    edge[tol].flow=flow;
    head[u]=tol++;
    edge[tol].from=v;
    edge[tol].to=u;
    edge[tol].next=head[v];
    edge[tol].flow=flow;
    head[v]=tol++;
} 

struct node {
    int v,w;
}cur,tail;
bool operator < (node a,node b) {
    return a.w>b.w;
}

int d[maxn];
int visit[maxn];
void dijkstra (int s,int t) {
    for (int i=0;i<maxn;i++) 
        d[i]=inf;
    memset(visit,0,sizeof(visit));
    d[s]=0;
    priority_queue<node> q;
    cur.v=s;
    cur.w=0;
    q.push(cur);
    while (!q.empty()) {
        cur=q.top();
        q.pop();
        int x=cur.v;
        if (visit[x]) continue;
        visit[x]=1;
        for (int i=head[x];i!=-1;i=edge[i].next) {
            if (d[edge[i].to]>d[x]+edge[i].flow) {
                d[edge[i].to]=d[x]+edge[i].flow;
                tail.v=edge[i].to;
                tail.w=d[edge[i].to];
                q.push(tail);
            }
        }
    }
    printf("%d\n",d[t]);
}


int main () {
    while (~scanf("%d%d",&N,&M)) {
        memset(head,-1,sizeof(head));
        tol=0;
        s=0;
        t=2*(N-1)*(M-1)+1;
        int x,y,cost;
        for (int i=1;i<=N;i++) {
            for (int j=1;j<M;j++) {
                scanf("%d",&cost);
                x=i==1?s:(2*(i-1)-1)*(M-1)+j;
                y=i==N?t:(2*(i-1))*(M-1)+j;
                addedge(x,y,cost);
            }
        }
        for (int i=1;i<N;i++) {
            for (int j=1;j<=M;j++) {
                scanf("%d",&cost);
                x=j==1?t:(2*(i-1))*(M-1)+j-1;
                y=j==M?s:(2*(i-1))*(M-1)+j-1+M;
                addedge(x,y,cost);
            }
        }
        for (int i=1;i<N;i++) {
            for (int j=1;j<M;j++) {
                scanf("%d",&cost);
                x=(2*(i-1))*(M-1)+j;
                y=(2*(i-1)+1)*(M-1)+j;
                addedge(x,y,cost); 
            }
        }
        dijkstra(s,t);
    }
    return 0;
}