SPOJ 287 Smart Network Administrator
2012-08-08 10:51
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SPOJ_287
YY了一下颜色的数量取决于网络流中各条边中最大的流量,于是二分边的容量并做网络流就可以了。
但至于为什么颜色的数量取决于网络流中各条边中最大的流量,暂时没有细加证明……
YY了一下颜色的数量取决于网络流中各条边中最大的流量,于是二分边的容量并做网络流就可以了。
但至于为什么颜色的数量取决于网络流中各条边中最大的流量,暂时没有细加证明……
#include<stdio.h> #include<string.h> #include<algorithm> #define MAXD 510 #define MAXM 501010 #define INF 0x3f3f3f3f int N, M, K, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM], list[MAXD]; int S, T, d[MAXD], q[MAXD], work[MAXD]; struct Edge { int x, y; }edge[MAXM]; void init() { int i; scanf("%d%d%d", &N, &M, &K); for(i = 0; i < K; i ++) scanf("%d", &list[i]); for(i = 0; i < M; i ++) scanf("%d%d", &edge[i].x, &edge[i].y); } void add(int x, int y, int z) { v[e] = y, flow[e] = z; next[e] = first[x], first[x] = e ++; } void build(int c) { int i; memset(first, -1, sizeof(first[0]) * (T + 1)); e = 0; add(S, 1, K), add(1, S, 0); for(i = 0; i < K; i ++) add(list[i], T, 1), add(T, list[i], 0); for(i = 0; i < M; i ++) add(edge[i].x, edge[i].y, c), add(edge[i].y, edge[i].x, c); } 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 min, max, mid; S = 0, T = N + 1; min = -1, max = K; for(;;) { mid = min + max + 1 >> 1; if(max == mid) break; build(mid); if(dinic() == K) max = mid; else min = mid; } printf("%d\n", mid); } int main() { int t; scanf("%d", &t); while(t --) { init(); solve(); } return 0; }
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