bzoj 1001 分类： bzoj 2015...-优快云博客

本文探讨了在信息学竞赛中解决平面图最大流问题的方法，通过构造对偶图并运用最短路径算法，实现复杂度为O(n*m*log²(n*m))的高效求解。代码示例详细展示了如何构建图模型及执行Dijkstra算法。

平面图最大流，看到数据范围我就怂了。

先构造平面图的对偶图，然后求最短路即可。
时间复杂度：O(n∗m∗log2(n∗m))

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <vector>
#include <utility>
#include <iostream>
#include <algorithm>

const int maxm = 1005, maxn = 1005;
const int maxnode = ((maxn-1)*(maxm-1)<<1);
const int maxe = maxnode * 3, INF = 1<<30;

struct Edge
{
    int v, w, next;
    Edge(){}
    Edge(int v,int w,int next):v(v),w(w),next(next){}
}edge[maxe];
int n, m, node, S, T;
int head[maxnode], el = 0; 

int read()
{
    int x = 0;char c = getchar();
    while(c < '0' || c > '9') {c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<3) + (x<<1) + (c-'0'); c = getchar();}
    return x;
}

void write(long long x)
{
    static char s[20];int sl = 0;
    while(x) s[++sl] = x%10 + '0',x /= 10;
    if(!sl) {putchar('0');return;}
    while(sl) putchar(s[sl--]);
}

void NewEdge(int u,int v,int w)
{
    edge[++el] = Edge(v, w, head[u]), head[u] = el;
}
void Build()
{
    int now, last, w;

    now = 2, last = 1 - ((m-1)<<1);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j < m; j++)
        {
            w = read();

            if(i == 1) 
                NewEdge(S, now, w), NewEdge(now, S, w);
            else if(i == n)
                NewEdge(last, T, w), NewEdge(T, last, w); 
            else
                NewEdge(last, now, w), NewEdge(now, last, w);

            now += 2, last += 2;    
        }
    now = 1, last = 0;      
    for(int i = 1; i < n; i++)
        for(int j = 1; j <= m; j++)
        {
            w = read();

            if(j == 1) 
                NewEdge(T, now, w), NewEdge(now, T, w);
            else if(j == m)
                NewEdge(last, S, w), NewEdge(S, last, w); 
            else
                NewEdge(last, now, w), NewEdge(now, last, w);

            if(j != m) now += 2, last += 2;
        }
    now = 2, last = 1;  
    for(int i = 1; i < n; i++)
        for(int j = 1; j < m; j++)
        {
            w = read();

            NewEdge(last, now, w), NewEdge(now, last, w);

            now += 2, last += 2;
        }
}

typedef std::pair<long long,int> Heapnode;
#define Root_Min std::vector<Heapnode>,std::greater<Heapnode> 
std::priority_queue<Heapnode, Root_Min> heap;
bool hash[maxnode];
long long dist[maxnode];

long long Dijstra()
{
    for(int i = 1; i <= node; i++) dist[i] = INF;

    dist[S] = 0, heap.push(std::make_pair(0, S));

    while(!heap.empty())
    {
        int t = heap.top().second; 
        heap.pop();

        if(hash[t]) continue;

        hash[t] = true;

        if(t == T) break;

        for(int i = head[t], p; i ; i = edge[i].next)
            if(!hash[p = edge[i].v])
            {
                long long tmp = dist[t] + edge[i].w;

                if(tmp < dist[p])
                    dist[p] = tmp,  heap.push(std::make_pair(tmp, p));
            }
    }

    return dist[T];
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj1001.in","r",stdin);
    freopen("bzoj1001.out","w",stdout);
#endif

    n = read(), m = read(), node = ((n-1)*(m-1)<<1) + 2;
    S = node - 1,  T = node;

    Build(), write(Dijstra());

#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);         
#endif
    return 0;   
}