The 11th chapter graph theory model and algorithm

Talk about trees again.

No root tree has a root number

#include <bits/stdc++.h>using namespace Std;const int maxn = 100;vector<int> g[maxn];int n;void read_tree () {    int u,v;    scanf ("%d", &n);        for (int i=0;i<n-1;i++) {scanf ("%d%d", &u,&v);        G[u].push_back (v);    G[v].push_back (U);    }}int p[100];void dfs (int u,int fa) {///Convert to U as root subtree, u parent node is FA int d=g[u].size ();        for (int i=0;i<d;i++) {int v=g[u][i];    if (V!=FA) DFS (V,P[V]=U);    }}int Main () {read_tree ();        for (int i=0;i<n-1;i++) {printf ("%d:", i);        for (int j=0;j<g[i].size (); j + +) printf ("%d", g[i][j]);    printf ("\ n");    } p[1]=-1;    DFS (1,-1); for (int i=0;i<n;i++) printf ("%d", I); printf ("\ n"); <pre name= "code" class= "CPP" >const int maxn = 100;int LCH[MAXN], RCH[MAXN]; Char OP[MAXN]; The left and right child node number and character int nc=0 of each node;    node int build_tree (char * s,int x,int y) {int i,c1=-1,c2=-1,p=0;    int u;        if (y-x==1) {//Enter a character to establish a separate node u = ++NC; lch[u]=rch[u]=0;        OP[U]=S[X]; return u;   } for (i=x;i<y;i++) {switch (S[i]) {case ' (':p ++;break;            Case ') ':p--;break; Case ' + ': Case '-': if (!p) c1=i;            Break Case ' * ': Case '/': if (!p) c2=i;        Break    }} if (c1<0) c1=c2;    if (c1<0) return Build_tree (s,x+1,y-1);    U=++NC;    Lch[u]=build_tree (S,X,C1);    Ech[u]=build_tree (S,c1+1,y);    OP[U]=S[C1]; return u;}

for (int i=0;i<n;i++) printf ("%d", p[i]); printf ("\ n"); /* Initial Tree 8 1 4 1 5 1 0 5 6 5 7 0 2 0 3 0:1 2 3 1:4 5 0 2:0 3:0 4:1 5:1 6 7 6:5 * return 0;}

An expression tree

const int MAXN = 100;int LCH[MAXN], RCH[MAXN]; Char OP[MAXN]; The left and right child node number and character int nc=0 of each node; node int build_tree (char * s,int x,int y) {    int i,c1=-1,c2=-1,p=0;    int u;    if (y-x==1) {//Enter a character to establish a separate node        u = ++NC;        lch[u]=rch[u]=0; OP[U]=S[X];        return u;    }    for (i=x;i<y;i++) {        switch (S[i]) {case            ' (':p ++;break;            Case ') ':p--;break;            Case ' + ': Case '-': if (!p) c1=i; break;            Case ' * ': Case '/': if (!p) c2=i; break;        }    }    if (c1<0) c1=c2;    if (c1<0) return Build_tree (s,x+1,y-1);    U=++NC;    Lch[u]=build_tree (S,X,C1);    Ech[u]=build_tree (s,c1+1,y);    OP[U]=S[C1];    return u;}

Example 11-1 common expression elimination uva12219

#include <bits/stdc++.h>using namespace Std;const int INF = 0x3f3f3f3f;const int MAXN = 105;struct Edge {int u,v,    Cost    BOOL operator < (const edge& a) const {return cost > a.cost; }};int f[maxn];int n,m;vector<edge> edges;int Find (int x) {return x==f[x]?x:f[x]=find (f[x]);}    int Kruskal (int k) {for (int i=0;i<=n;i++) f[i]=i;    int cnt=0;    int minn=inf,maxn=0;        for (int i=k;i<m;i++) {Edge e = edges[i];        int v=e.v;        int u=e.u;        int fu=find (u);        int Fv=find (v);            if (FU!=FV) {cnt++;            F[FU]=FV;            Minn=min (E.cost,minn);        MAXN =max (E.COST,MAXN);    }} if (cnt!=n-1) return-1; else return Maxn-minn;}        int main () {while (~scanf ("%d%d", &n,&m)) {if (n==0&&m==0) break;        Edges.clear ();            for (int i=1;i<=m;i++) {int a,b,c;            scanf ("%d%d%d", &a,&b,&c); Edges.push_back (Edge) {a,b,c});        } sort (Edges.begin (), Edges.end ());        int ans=inf;            for (int i=0;i<=m;i++) {int T=kruskal (i);            if (t==-1) break;        ans = min (ans,t);        } if (ans = = INF) printf ("%d\n",-1);    else printf ("%d\n", ans); } return 0;}

Example 11-3 buy or Build uva1151

#include <bits/stdc++.h>using namespace Std;const int maxn = 1005;struct point {int x, y;} pp[maxn];struct edge{int s,e,dist;} L[maxn*maxn];int n,q,m;int p[maxn];vector<int> g[10];int c[10];int distance_ (point A,point B) {return (a.x-b.x) * ( a.x-b.x) + (A.Y-B.Y) * (A.Y-B.Y);} BOOL CMP (Edge A,edge b) {return a.dist < b.dist;} int find_ (int x) {return p[x]==x?x:p[x]=find_ (p[x]);}    BOOL Merge_ (int a,int b) {int x=find_ (a);    int Y=find_ (b);    if (x==y) return false; P[x]=y; return true;}    int Kruskal () {int ans=0;    int num=0;            for (int i=0;i<m&&num<n-1;i++) {if (Merge_ (L[I].S,L[I].E)) {num++;        Ans + = l[i].dist; }} return ans;    void Solve () {for (int i=0;i<=n;i++) p[i]=i;    int Ans=kruskal ();        for (int s=1;s< (1&LT;&LT;Q); s++) {int cost=0;        for (int tt=0;tt<=n;tt++) P[tt]=tt; for (int j=0;j<q;j++) {if (!) (            (S&GT;&GT;J) &1) continue; Cost + =C[J];            for (int k=0;k<g[j].size (); k++) {Merge_ (g[j][k],g[j][0]);    }} ans =min (ans, Cost+kruskal ()); } printf ("%d\n", ans);}    int main () {int t;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&q);        for (int i=0;i<10;i++) g[i].clear ();            for (int i=0;i<q;i++) {int cnt;            scanf ("%d%d", &cnt,&c[i]);            int A;                for (int j=0;j<cnt;j++) {scanf ("%d", &a);            G[i].push_back (a);        }} for (int i=1;i<=n;i++) {scanf ("%d%d", &pp[i].x,&pp[i].y);        } m=0;                for (int i=1;i<=n;i++) {for (int j=i+1;j<=n;j++) {l[m].s=i;                L[m].e=j;            L[m++].dist=distance_ (Pp[i],pp[j]);        }} sort (l,l+m,cmp);        Solve ();    if (t) printf ("\ n"); } return 0;}

Dijkstra algorithm:

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The 11th chapter graph theory model and algorithm