UVA 1376 Animal Run 最短路

平面图最小割转最短路

书上是说以边为结点建图

但是我觉得好像以每块空白区域为结点建图会更自然点。把矩形的右上方编号为0，左下方编号为1，分别为起点终点

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("in.txt","r",stdin);
    //freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}
struct HeapNode
{
    int d,u;
    bool operator < (const HeapNode &ant) const
    {
        return d>ant.d;
    }
};
struct Edge
{
    int from,to;
    int dist;
};
const int maxn=2000005;
struct Dijksta
{
    int n;
    vector<int> g[maxn];
    vector<Edge> edge;
    int done[maxn];
    int d[maxn];

    void init(int n)
    {
        this->n=n;
        for(int i=0;i<=n;i++)
            g[i].clear();
        edge.clear();
    }

    void add(int u,int v,int w)
    {
        Edge e=(Edge){u,v,w};
        edge.push_back(e);
        g[u].push_back(edge.size()-1);
    }

    void solve(int s)
    {
        for(int i=0;i<=n;i++)
            d[i]=INF;
        memset(done,0,sizeof(done));
        d[s]=0;
        priority_queue<HeapNode> q;
        q.push((HeapNode){0,s});
        while(!q.empty())
        {
            HeapNode x=q.top();q.pop();
            if(done[x.u])continue;
            int u=x.u;
            done[u]=1;
            for(int i=0;i<g[u].size();i++)
            {
                Edge &e=edge[g[u][i]];
                if(d[u]+e.dist<d[e.to])
                {
                    d[e.to]=d[u]+e.dist;
                    q.push((HeapNode){d[e.to],e.to});
                }
            }
        }
    }
};

Dijksta solver;
int n,m;
int id[1005][1005][2];
const int UP=0,DOWN=1;
int vcnt;
int ID(int a,int b,int d)
{
    if(a<1||b>=m)return 0;
    if(a>=n||b<1)return 1;
    int &x=id[a][b][d];
    if(x==0)x=++vcnt;
    return x;
}

int main()
{
    //freopen("d:\in1.txt","r",stdin);
    int ca=0;
    while(scanf("%d%d",&n,&m)!=EOF&&m)
    {
        solver.init(n*m*2+2);
        vcnt=2;
        memset(id,0,sizeof(id));
        //horizontal
        for(int i=1;i<=n;i++)
            for(int j=1;j<m;j++)
            {
                int u,v,w;
                scanf("%d",&w);
                u=ID(i-1,j,DOWN);
                v=ID(i,j,UP);
                solver.add(u,v,w);
                solver.add(v,u,w);
            }

        //vertical
        for(int i=1;i<n;i++)
            for(int j=1;j<=m;j++)
            {
                int u,v,w;
                scanf("%d",&w);
                u=ID(i,j-1,UP);
                v=ID(i,j,DOWN);
                solver.add(u,v,w);
                solver.add(v,u,w);
            }
        //diagonal
        for(int i=1;i<n;i++)
            for(int j=1;j<m;j++)
            {
                int u,v,w;
                scanf("%d",&w);
                u=ID(i,j,UP);
                v=ID(i,j,DOWN);
                solver.add(u,v,w);
                solver.add(v,u,w);
            }
        solver.solve(0);
        printf("Case %d: Minimum = %d
",++ca,solver.d[1]);
    }
    return 0;
}