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

【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;
}