【POJ1637】Sightseeing tour【最大流】【混合图欧拉回路】

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【题目链接】

混合图欧拉回路。

论文题，见【网络流建模汇总】

注意入度大于出度必须连接汇点，出度大于入度必须源点去连。不能相反。

/* Pigonometry */#include <cstdio>#include <algorithm>using namespace std;const int maxn = 205, maxm = 5005, maxq = 10000, inf = 0x3f3f3f3f;int n, m, cur[maxn], head[maxn], cnt, depth[maxn], bg, ed, q[maxq], in[maxn], out[maxn];struct _edge {int v, w, next;} g[maxm];inline int iread() {int f = 1, x = 0; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';return f * x;}inline void add(int u, int v, int w) {g[cnt] = (_edge){v, w, head[u]};head[u] = cnt++;}inline bool bfs() {for(int i = 0; i <= ed; i++) depth[i] = -1;int h = 0, t = 0, u, i; depth[q[t++] = bg] = 1;while(h != t) for(i = head[u = q[h++]]; ~i; i = g[i].next) if(g[i].w && !~depth[g[i].v]) {depth[g[i].v] = depth[u] + 1;if(g[i].v == ed) return 1;q[t++] = g[i].v;}return 0;}inline int dfs(int x, int flow) {if(x == ed) return flow;int left = flow;for(int i = cur[x]; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] == depth[x] + 1) {int tmp = dfs(g[i].v, min(left, g[i].w));left -= tmp; g[i].w -= tmp; g[i ^ 1].w += tmp;if(g[i].w) cur[x] = i;if(!left) return flow;}if(left == flow) depth[x] = -1;return flow - left;} int main() {for(int T = iread(); T; T--) {n = iread(); m = iread(); bg = 0; ed = n + 1;for(int i = 0; i <= ed; i++) in[i] = out[i] = 0, head[i] = -1; cnt = 0;for(int i = 1; i <= m; i++) {int u = iread(), v = iread(), opt = iread();out[u]++; in[v]++;if(!opt) add(u, v, 1), add(v, u, 0);}int ans = 0;bool flag = 0;for(int i = 1; i <= n; i++) {int w = in[i] - out[i];if(w & 1) {flag = 1;break;}if(w < 0) w = -w, w >>= 1, ans += w, add(bg, i, w), add(i, bg, 0);else if(w > 0) w >>= 1, add(i, ed, w), add(ed, i, 0);}if(flag) {printf("impossible\n");continue;}while(bfs()) {for(int i = 0; i <= ed; i++) cur[i] = head[i];ans -= dfs(bg, inf);}if(ans) printf("im");printf("possible\n");}return 0;}

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