平面图最小割 对偶图:
平面图G的性质:
(1)满足n个点,m条边,f个面 f = m - n + 2;
(2)存在与其对应的对偶图G*;
对偶图:将原图中每个面变成一个点，外边界的无限大的面看成一个点，后连线即成对偶图；
G的面数等于G*的点数，边数相等；
详解请看 最大最小定理(平面图最小割 对偶图)周冬
对于平面图的最大流(最小割)只需转化为对偶图，直接跑最短路即可；
ps：觉得建图是最复杂的，各种RE(边数就是原来的边数，只是点数变成了原来的面数,,不注意就RE了)；还有现在行数列数都变成减了1；原来的横线现在也变成了竖线；耐心点..一个debug的方法就是在ins插入边时输出，这样觉得挺好；
对偶图裸题:
// 296MS 12452K
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<stack>
#include<set>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define inf 0x3f3f3f3f
typedef __int64 ll;
template<typename T>
void read1(T &m)
{
T x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
if(a>9) out(a/10);
putchar(a%10+‘0‘);
}
const int M = 404*404*2;
int head[M<<1],tot;
struct Edge{
int to,w,Next;
Edge(){}
Edge(int to,int w,int nx):to(to),w(w),Next(nx){}
}e[M<<1];
inline void ins(int u,int v,int w)
{
e[tot] = Edge{v,w,head[u]};
head[u] = tot++;
}
typedef pair<int,int> PII;
#define MK make_pair
#define A first
#define B second
priority_queue<PII,vector<PII>,greater<PII> > Q;
bool vs[160016];
int dist[M];
ll Djistra(int s,int t)
{
while(!Q.empty()) Q.pop();
fill(vs,vs+t+1,false);
fill(dist,dist+t+1,inf);
dist[s] = 0;
Q.push(MK(0,s));
while(!Q.empty()){
PII tmp = Q.top();Q.pop();
int u = tmp.B;
if(vs[u]) continue;
vs[u] = true;
if(u == t) return dist[t];
for(int id = head[u];~id;id = e[id].Next){
int v = e[id].to,cost = e[id].w;
if(dist[v] > dist[u] + cost){
dist[v] = dist[u] + cost;
Q.push(MK(dist[v],v));
}
}
}
}
int main()
{
int n,T,kase = 1;
read1(T);
while(T--){
read1(n);
int s = 0, t = (n-1)*(n-1)+1;
int u,v1,v2,x;
MS1(head);tot = 0;
rep0(i,0,n-1){ // n-1行特殊，只要加竖线即可;
rep0(j,1,n){//注意j一定要从1开始;
read1(x);
u = i*(n-1)+j;//一般情况只需要找到当前节点的上面v1和左边v2
v1 = (i-1)*(n-1)+j;
v2 = i*(n-1)+j-1;
if(i == 0) v1 = s;
if(j == 1) v2 = t;
ins(u,v1,x);ins(v1,u,x);
ins(u,v2,x);ins(v2,u,x);
}
read1(x);
u = i*(n-1)+n-1;
ins(u,s,x);ins(s,u,x);
}
rep0(j,1,n){
read1(x);
u = (n-2)*(n-1)+j;
ins(u,t,x);ins(t,u,x);
}
read1(x);
out(Djistra(s,t));
puts("");
}
return 0;
}
前面使用优化的Dinic算法，用了1984ms,不是Dinic太慢(这还是较快的网络流算法了) 这道题原本就需要模型转化，变成最短路使用Djistra+优先队列 348ms;
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<stack>
#include<set>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define inf 0x3f3f3f3f
template<typename T>
void read1(T &m)
{
T x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
if(a>9) out(a/10);
putchar(a%10+‘0‘);
}
const int M = 1004*1004*2;
int head[M*3],tot;
struct Edge{
int to,w,Next;
Edge(){}
Edge(int to,int w,int Next):to(to),w(w),Next(Next){}
}e[M*3];
inline void ins(int u,int v,int w)
{
//cout<<" ........... "<<u<<" "<<v<<" "<<w<<endl;
e[tot] = Edge{v,w,head[u]};
head[u] = tot++;
}
typedef pair<int,int> PII;
#define MK make_pair
#define A first
#define B second
priority_queue<PII,vector<PII>,greater<PII> > Q;
bool vs[M];
int dist[M];
int Djistra(int s,int t)
{
while(!Q.empty()) Q.pop();
fill(vs,vs+t+1,false);
fill(dist,dist+t+1,inf);
dist[s] = 0;
Q.push(MK(0,s));
while(!Q.empty()){
PII tmp = Q.top();Q.pop();
int u = tmp.B;
if(vs[u]) continue;
vs[u] = true;
if(u == t) return dist[t];
for(int id = head[u];~id;id = e[id].Next){
int v = e[id].to,cost = e[id].w;
if(dist[v] > dist[u] + cost){
dist[v] = dist[u] + cost;
Q.push(MK(dist[v],v));
}
}
}
}
int main()
{
int n,m,x;
read2(n,m);
int s = 0,t = (n-1)*(m-1)*2+1;
if(n == 1 || m == 1){//坑点不能建对偶图，没意思
int ans = inf;
if(n != 1) swap(n,m);
rep0(i,1,m){
read1(x);
ans = min(ans,x);
}
return out(ans == inf?0:ans),0;
}
fill(head,head+t+1,-1);tot = 0;
rep0(i,0,n){
rep0(j,1,m){
read1(x);
int u = (i*(m-1)+j)*2,
v = ((i-1)*(m-1)+j)*2-1;
if(i == 0) v = s;
if(i == n-1) u =u-2*(m-1)-1, v = t;
ins(u,v,x);ins(v,u,x);
}
}
rep0(i,0,n-1){
rep1(j,1,m){
read1(x);
int u = (i*(m-1)+j)*2-1,
v = u-1;
if(j == 1) v = t;
if(j == m) u -= 1,v = s;
ins(u,v,x);ins(v,u,x);
}
}
rep0(i,0,n-1){
rep0(j,1,m){
read1(x);
int u = (i*(m-1)+j)*2,
v = u - 1;
ins(u,v,x);ins(v,u,x);
}
}
out(Djistra(s,t));
return 0;
}
时间： 2024-12-27 14:02:15