UVALive 3661 最小割转化为最短路径

题意：动物逃跑，从左上跑到右下角，逃跑的路径是一个grid的边，现在动物园的工作人员要去拦截。input给出了grid每条路径拦截所需要的cost，题目要求拦截成功最小的成本。

经典的最小割问题，但是400*400个点太大了，所以不能直接这么做

lrj给出的方法是动物要从左上跑到右下，所有我们考虑怎么摆障碍物就可以了，思考可以知道，只要做左/下->右/上的连续拦截边，就可以将所有可行路径分割开。如下图(偷了别人博客的图。。。)

这张图其实是hdu3870的图，题意差不多，但是没有斜边，所以少了两倍的点，建图也简单的多。从左/下到右/上的拦截边可以这样考虑，增加s源点，ed终点，最在要求拦截的成功的最小成本就是从s到t可最短可行边，dij模板套一下就可以了

难点在思考和建图，边转化为点，重新建图刚开始看还是挺毁我三观的

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <queue>
#include <vector>
#include <algorithm>
#include <ctime>
using namespace std;

#define EdsonLin

#ifdef EdsonLin
#define debug(...) fprintf(stderr,__VA_ARGS__)
#else
#define debug(...)
#endif //EdsonLin

typedef long long ll;
typedef double db;
const int inf = 0x3f3f3f;
const int MAXN = 1e3;
const int MAXNN = 2e6+100;
//const int MAXM = 1e6;
//const int MAXM = 3e4+100;
const int MOD = 1000000007;
const db eps = 1e-3;
#define PB push_back
#define UP(i) ((i)<<1)
#define DOWN(i) ((UP(i))-1)

struct dij{
    int n,m;
    int first[MAXNN];
    struct edge{
        int st,to,next,dist;
    };
    vector<edge>e;
    int top;
    int d[MAXNN]; //s到各节点的距离
    int done[MAXNN]; //是否已经被永久标记
    int p[MAXNN]; //记录上一条边
    struct heapnode{
        int st;
        int dist;
        bool operator < (const heapnode& rhs) const {
            return dist>rhs.dist;
        }
    };
    void init(int n){
        this->n = n;
        memset(first,-1,sizeof(first));
        top = 0;
        e.clear();
    }
    void addedge(int u,int v,int dist){
        /*e[top].st = u;
        e[top].to = v;
        e[top].dist = dist;
        e[top].next = first[u];
        first[u] = top++;*/
        edge a;
        a.st = u;
        a.to = v;
        a.dist = dist;
        a.next  = first[u];
        e.PB(a);
        first[u] = top++;
        //cout<<first[u]<<endl;
        //cout<<top<<endl;
    }

    void pqdij(int s){
        priority_queue<heapnode>Q;
        heapnode a;
        for(int i=0;i<n;i++)
            d[i] = inf;
        d[s] = 0;
        memset(done,0,sizeof(done));
        a.dist = 0;
        a.st = s;
        Q.push(a);
        while(!Q.empty()){
            heapnode x = Q.top();
            Q.pop();
            int u = x.st;
            if(done[u])continue;
            done[u] = 1;
            for(int i=first[u];i!=-1;i=e[i].next){
                if(d[e[i].to]>d[u]+e[i].dist){
                    d[e[i].to] = d[u] + e[i].dist;
                    p[e[i].to] = i;
                    a.dist = d[e[i].to];
                    a.st = e[i].to;
                    Q.push(a);
                }
            }
        }
    }
}solver;

int hor[MAXN][MAXN],vec[MAXN][MAXN],dia[MAXN][MAXN];
int readint(){int x;scanf("%d",&x);return x;}

int main()
{
    #ifdef EdsonLin
        //freopen("1.in","r",stdin);
        //freopen("1.out","w",stdout);
        int _time_ed = clock();
    #endif //EdsonLin
    int mc=0,n,m;
    while(scanf("%d%d",&n,&m)&&n){
        for(int i=1;i<=n;i++){
            for(int j=1;j<m;j++)
                hor[i][j] = readint();
        }
        for(int i=1;i<n;i++){
            for(int j=1;j<=m;j++)
                vec[i][j] = readint();
        }
        for(int i=1;i<n;i++){
            for(int j=1;j<m;j++)
                dia[i][j] = readint();
        }
        int st=0,ed = (n-1)*(m-1)*2+1;
        solver.init(ed+1);
        for(int i=1;i<n;i++){
            for(int j=1;j<m;j++){
                if(i==1){
                    solver.addedge(2*(i-1)*(m-1)+UP(j),ed,hor[i][j]);
                    solver.addedge(ed,2*(i-1)*(m-1)+UP(j),hor[i][j]);
                }
                if(i==n-1){
                    solver.addedge(st,2*(i-1)*(m-1)+DOWN(j),hor[i+1][j]);
                    solver.addedge(2*(i-1)*(m-1)+DOWN(j),st,hor[i+1][j]);
                }
                if(i!=n-1){
                    solver.addedge(2*(i-1)*(m-1)+DOWN(j),2*(i)*(m-1)+UP(j),hor[i+1][j]);
                    solver.addedge(2*i*(m-1)+UP(j),2*(i-1)*(m-1)+DOWN(j),hor[i+1][j]);
                }
                if(j==1){
                    solver.addedge(st,2*(i-1)*(m-1)+DOWN(j),vec[i][j]);
                    solver.addedge(2*(i-1)*(m-1)+DOWN(j),st,vec[i][j]);
                }
                if(j!=m-1){
                    solver.addedge(2*(i-1)*(m-1)+DOWN(j),2*(i-1)*(m-1)+UP(j),dia[i][j]);
                    solver.addedge(2*(i-1)*(m-1)+UP(j),2*(i-1)*(m-1)+DOWN(j),dia[i][j]);

                    solver.addedge(2*(i-1)*(m-1)+UP(j),2*(i-1)*(m-1)+DOWN(j+1),vec[i][j+1]);
                    solver.addedge(2*(i-1)*(m-1)+DOWN(j+1),2*(i-1)*(m-1)+UP(j),vec[i][j+1]);
                }
                if(j==m-1){
                    solver.addedge(2*(i-1)*(m-1)+DOWN(j),2*(i-1)*(m-1)+UP(j),dia[i][j]);
                    solver.addedge(2*(i-1)*(m-1)+UP(j),2*(i-1)*(m-1)+DOWN(j),dia[i][j]);

                    solver.addedge(2*(i-1)*(m-1)+UP(j),ed,vec[i][j+1]);
                    solver.addedge(ed,2*(i-1)*(m-1)+UP(j),vec[i][j+1]);
                }
            }
        }
        solver.pqdij(st);
        printf("Case %d: Minimum = %d
",++mc,solver.d[ed]);
    }

    #ifdef EdsonLin
        debug("time: %d
",int(clock()-_time_ed));
    #endif //EdsonLin
   // cout << "Hello world!" << endl;
    return 0;
}