SGU_326
这个题目的武大OJ的1124基本是一样的，只不过多了队外的比赛，对于队外的比赛只要让第1组全部赢而其他组全部输即可，剩下的就和WOJ_1124是一样的处理方式了，具体的思路可以参考我的WOJ_1124的题解：。
#include#include #include #define MAXN 25#define MAXD 430#define MAXM 2450#define INF 0x3f3f3f3fint N, SUM, score[MAXN], r[MAXN], first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM];int S, T, d[MAXD], q[MAXD], work[MAXD];void add(int x, int y, int z){ v[e] = y, flow[e] = z; next[e] = first[x], first[x] = e ++;}void init(){ int i, j, a; for(i = 1; i <= N; i ++) scanf("%d", &score[i]); for(i = 1; i <= N; i ++) scanf("%d", &r[i]); score[1] += r[1]; S = 0, T = N + N * N + 1; memset(first, -1, sizeof(first[0]) * (T + 1)); e = SUM = 0; for(i = 1; i <= N; i ++) for(j = 1; j <= N; j ++) { scanf("%d", &a); if(a && i < j && i != 1) { SUM += a; add(i, i * N + j, a), add(i * N + j, i, 0); add(j, i * N + j, a), add(i * N + j, j, 0); add(i * N + j, T, a), add(T, i * N + j, 0); } }}int bfs(){ int i, j, rear = 0; memset(d, -1, sizeof(d[0]) * (T + 1)); d[S] = 0, q[rear ++] = S; for(i = 0; i < rear; i ++) for(j = first[q[i]]; j != -1; j = next[j]) if(flow[j] && d[v[j]] == -1) { d[v[j]] = d[q[i]] + 1, q[rear ++] = v[j]; if(v[j] == T) return 1; } return 0;}int dfs(int cur, int a){ if(cur == T) return a; int t; for(int &i = work[cur]; i != -1; i = next[i]) if(flow[i] && d[v[i]] == d[cur] + 1) if(t = dfs(v[i], std::min(a, flow[i]))) { flow[i] -= t, flow[i ^ 1] += t; return t; } return 0;}int dinic(){ int ans = 0, t; while(bfs()) { memcpy(work, first, sizeof(first[0]) * (T + 1)); while(t = dfs(S, INF)) ans += t; } return ans;}void solve(){ int i, j, k; for(i = 2; i <= N; i ++) if(score[i] > score[1]) { printf("NO\n"); return ; } for(i = 2; i <= N; i ++) add(S, i, score[1] - score[i]), add(i, S, 0); printf("%s\n", dinic() == SUM ? "YES" : "NO");}int main(){ while(scanf("%d", &N) == 1) { init(); solve(); } return 0; }