Uva 11506 Angry Programmer(最小分割最大流、分割点、建図)

この問題は、初めて表に隣接して書いたので、waして、それから行列を変えて、結果はやはり間違っていましたが、どこが間違っているのか気づきました.
主に図を建てて、2つの点が起点か送金点かどうかを判断して、この2つの点に対して、分解しないでください、費用がないためです
残りのすべての点は、2つの点に分解して、容量はこのcomputerの費用で、その他は規則に従って、元の点から入って、分解した点から出ます.
行列のコードは次のとおりです.

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int N = 100;
const int INF = 100000000;
int n, m;
int cap[N][N], flow[N][N];
int p[N], a[N];

int maxflow() {
    queue<int> q;
    memset(flow, 0, sizeof(flow));
    int f = 0;
    while ( 1 ) {
        memset(a, 0, sizeof(a));
        a[1] = INF;
        q.push(1);
        while ( !q.empty() ) {
            int u = q.front(); q.pop();
            for ( int v = 1; v <= n + n - 1; v++ ) if ( !a[v] && cap[u][v] > flow[u][v] ) {
                p[v] = u;
                a[v] = min( a[u], cap[u][v] - flow[u][v] );
                q.push(v);
            }
        }
        if ( a[n] == 0 ) break;
        for ( int u = n; u != 1; u = p[u] ) {
            //printf("%d ", u );
            flow[p[u]][u] += a[n];
            flow[u][p[u]] -= a[n];
        }
        f += a[n];
    }
    return f;
}
int main()
{
    while ( scanf("%d%d", &n, &m) != EOF && !( !n && !m ) ) {
        memset( cap, 0, sizeof(cap) );
        for ( int i = 2; i < n; ++i ) {
            int com, c;
            scanf("%d%d", &com, &c);
            cap[com][com+n] = cap[com+n][com] = c;
        }
        for ( int i = 0; i < m; ++i ) {
            int u, v, c;
            scanf("%d%d%d", &u, &v, &c);
            if ( u == 1 && v != n ) cap[u][v] = cap[v+n][u] = c;
            else if ( u == 1 && v == n ) cap[u][v] = cap[v][u] = c;
            else if ( u == n && v != 1 ) cap[v+n][u] = cap[u][v] = c;
            else if ( u == n && v == 1 ) cap[u][v] = cap[v][u] = c;
            else cap[u+n][v] = cap[v+n][u] = c;
        }
        /*for ( int i = 1; i < 2*n; ++i ) { 
            for ( int j = 1; j < 2*n; ++j ) printf("%d ", cap[i][j]);
            printf("
");
        }*/
        int ans = maxflow();
        printf("%d
", ans);
    }
}

隣接テーブルコード:(辺数に注意×4)

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int N = 150;
const int M = 4500;
const int INF = 100000000;
struct edge {
    int v, next, cap, flow;
}e[M];
int n, m, id;
int head[N];

int maxflow() {
    int f = 0, a[N], p[N], r[N];
    while ( 1 ) {
        queue <int> q;
        int s = 1, t = n, np = n + n - 1;
        q.push(s);
        memset( a, 0, sizeof(a) );
        a[s] = INF;
        while ( !q.empty() ) {
            int u = q.front(); q.pop();
            for ( int i = head[u]; i != -1; i = e[i].next ) {
                int v = e[i].v, cap = e[i].cap, flow = e[i].flow;
                if ( cap > flow && !a[v] ) {
                    a[v] = min( a[u], cap - flow );
                    p[v] = u;
                    r[v] = i;
                    q.push(v);
                }
            }
        }
        if ( a[t] == 0 ) break;
        for ( int u = t; u != s; u = p[u] ) {
            int ei = r[u];
            e[ei].flow += a[t];
            e[ei^1].flow -= a[t];
        }
        f += a[t];
    }
    return f;
}
void add( int u, int v, int c ) {
    e[id].next = head[u];
    e[id].v = v, e[id].cap = c, e[id].flow = 0;
    head[u] = id++;

    e[id].next = head[v];
    e[id].v = u, e[id].cap = 0, e[id].flow = 0;
    head[v] = id++;
}
int main()
{
    while ( scanf("%d%d", &n, &m) != EOF && !(!n && !m) ) {
        id = 0;
        for ( int i = 0; i <= 2*n; ++i ) head[i] = -1;
        for ( int i = 2, node, cost; i < n; ++i ) {
          scanf("%d%d", &node, &cost); 
          add( node, node + n, cost );
          add( node + n, node, cost );
        }
        for ( int i = 0; i < m; ++i ) {
            int u, v, c;
            scanf("%d%d%d", &u, &v, &c);
            if ( u == 1 && v == n )  add( u, v, c ), add( v, u, c );
            else if ( u == 1 && v != n ) add( u, v, c ), add( v+n, u, c ) ;
            else if ( v == 1 && u == n ) add( v, u, c ), add( u, v, c );
            else add( v + n, u, c ), add( u+n, v, c );
        }
        printf("%d
", maxflow() );
    }
}