平面图最小割 对偶图
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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; }
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