首先是n^2m的复杂度的最小割 果断超时了 。。网上一搜是需要转换成最短路的
#include<iostream>
#include<cstring>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std;
const int N = 410 * 410,M = N * 2;
int head[N],to[M],last[M],c[M],cnt;
void add(int a,int b,int c1){
to[++cnt] = b;
c[cnt] = c1;
last[cnt] = head[a];
head[a] = cnt;
to[++cnt] = a;
c[cnt] = 0;
last[cnt] = head[b];
head[b] = cnt;
}
int S,T;int d[N],cur[N];
bool bfs(){
memset(d,0,sizeof d);
d[S] = 1;
queue<int>q;
q.push(S);cur[S] = head[S];
while(q.size()){
int p = q.front();
q.pop();
for(int i = head[p]; i != -1; i = last[i]){
int j = to[i];
if(!d[j] && c[i]){
d[j] = d[p] + 1;
q.push(j);
cur[j] = head[j];
if(j == T) return true;
}
}
}
return d[T] > 0;
}
int dfs(int x,int sum){
if(x == T) return sum;
int used = 0;
for(int i = cur[x]; i != -1; i = last[i]){
int j = to[i];
cur[x] = i;
if(d[j] == d[x] + 1 && c[i]){
int dd = dfs(j,min(sum - used,c[i]));
c[i] -= dd;
c[i ^ 1] += dd;
used += dd;
}
}
if(used == 0) d[x] = -1;
return used;
}
int dinic(){
int sum = 0;
while(bfs()){
sum += dfs(S,INF);
}
return sum;
}
int main(){
int n;
cin >> n;
memset(head,-1,sizeof head);
cnt = 1;
S = 0,T = n * n + 1;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
int x;
scanf("%d",&x);
if(i + 1 <= n) add((i - 1) * n + j,i * n + j,x);
if(j + 1 <= n) add((i - 1) * n + j,(i - 1) * n + j + 1,x);
}
}
add(S,1,INF);
add(n * n,T,INF);
cout << dinic() << endl;
return 0;
}
如何转换成求最短路的呢 每个平面图都对应一个图 每一个S -> T 的路径 都是一条割
#include<iostream>
#include<cstring>
#include<queue>
#include<utility>
#define x first
#define y second
#define INF 0x3f3f3f3f
using namespace std;
const int N = 410 * 410,M = N * 2;
typedef pair<int,int> PII;
int head[N],to[M],last[M],w[M],cnt;
void add(int a,int b,int c){
to[++cnt] = b;
w[cnt] = c;
last[cnt] = head[a];
head[a] = cnt;
}
int a[410][410];
int S,T;
int dist[N],flag[N];
int dij(){
memset(dist,0x3f,sizeof dist);
memset(flag,0,sizeof flag);
priority_queue<PII,vector<PII>,greater<PII> > q;
q.push({
0,S});
dist[S] = 0;
while(q.size()){
PII p = q.top();
q.pop();
if(flag[p.y]) continue;
flag[p.y] = 1;
for(int i = head[p.y]; i != -1; i = last[i]){
int j = to[i];
if(dist[j] > dist[p.y] + w[i]){
dist[j] = dist[p.y] + w[i];
q.push({
dist[j],j});
}
}
}
return dist[T];
}
int main(){
int n;
cin >> n;
memset(head,-1,sizeof head);
cnt = 1;
S = 0,T = n * n + 1;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
scanf("%d",&a[i][j]);
}
}
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
if(j == 1 && i != 1){
add((i - 2) * n + j + 1,T,a[i - 1][j]);
}
if(i == n && j != n){
add((i - 2) * n + j + 1,T,a[i][j]);
}
if(i == 1 && j != 1){
add(S,(i - 1) * n + j,a[i][j - 1]);
}
if(j == n && i != n){
add(S,(i - 1) * n + j,a[i][j]);
}
if(i + 1 < n){
add((i - 1) * n + j,i * n + j,a[i + 1][j - 1]);
}
if(j - 1 > 1){
add((i - 1) * n + j,(i - 1) * n + j - 1,a[i][j - 1]);
}
}
}
cout << dij();
return 0;
}