可行流应该是跑完 Dinic 之后的从 $T$ 到 $S$ 的反向边的剩余容量，而不是这次跑 Dinic 跑出来的流。但是，就算按照这种错误方法写，仍然可以通过此题。

错误代码

#define _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include <cstring>
#include <queue>
#define int long long
using namespace std;

const int N = 370, M = 1e3 + 10, C = 3e2 + 10, INF = 0x3f3f3f3f;

int n, m, s, t, ss, tt;
struct node {
	int from, to, next, siz;
	node() {}
	node(int _from, int _to, int _next, int _siz) {
		from = _from, to = _to, next = _next, siz = _siz;
	}
}ed[(N * C + M) << 1];
int pre[N + M], rad[N + M];

void add(int x, int y, int siz) {
	ed[++pre[0]] = node(x, y, pre[x], siz);
	pre[x] = pre[0];
	ed[++pre[0]] = node(y, x, pre[y], 0);
	pre[y] = pre[0];
}

int tflow[N + M];

void clear() {
	memset(tflow, 0, sizeof(tflow));
	memset(pre, 0, sizeof(pre));
	pre[0] = 1;
}

int dep[N + M];
queue<int> q;

bool bfs() {
	memset(dep, 0, sizeof(dep));
	dep[s] = 1;
	q.push(s);
	while (!q.empty()) {
		int now = q.front();
		q.pop();
		rad[now] = pre[now];
		for (int i = pre[now]; i != 0; i = ed[i].next) {
			if (dep[ed[i].to] != 0 || ed[i].siz == 0) continue;
			dep[ed[i].to] = dep[now] + 1;
			q.push(ed[i].to);
		}
	}
	return dep[t] != 0;
}

int dfs(int now, int sum) {
	if (now == t) return sum;
	int val = sum;
	for (int i = rad[now]; i != 0; i = ed[i].next) {
		rad[now] = i;
		if (dep[ed[i].to] != dep[now] + 1 || ed[i].siz == 0) continue;
		int tmp = dfs(ed[i].to, min(val, ed[i].siz));
		ed[i].siz -= tmp;
		ed[i ^ 1].siz += tmp;
		val -= tmp;
		if (val == 0) break;
	}
	return sum - val;
}

int dinic() {
	int res = 0;
	while (bfs()) {
		res += dfs(s, INF);
	}
	return res;
}

signed main() {
	while (scanf("%lld%lld", &n, &m) != EOF) {
		clear();
		s = n + m + 1, t = n + m + 2, ss = n + m + 3, tt = n + m + 4;
		for (int i = 1; i <= m; i++) {
			int g;
			scanf("%lld", &g);
			add(n + i, t, INF);
			tflow[n + i] -= g;
			tflow[t] += g;
		}
		for (int i = 1; i <= n; i++) {
			int c, d;
			scanf("%lld%lld", &c, &d);
			add(s, i, d);
			for (int j = 1; j <= c; j++) {
				int _t, l, r;
				scanf("%lld%lld%lld", &_t, &l, &r);
				_t++;
				add(i, n + _t, r - l);
				tflow[i] -= l;
				tflow[n + _t] += l;
			}
		}
		int mxflow = 0;
		for (int i = 1; i <= t; i++)
			if (tflow[i] < 0) add(i, tt, -tflow[i]);
			else if (tflow[i] > 0) add(ss, i, tflow[i]), mxflow += tflow[i];
		add(t, s, INF);
		s = ss, t = tt;
		if (dinic() != mxflow) {
			printf("-1\n\n");
			continue;
		}
		ed[pre[0]].siz = ed[pre[0] ^ 1].siz = 0;
		s = n + m + 1, t = n + m + 2;
		printf("%d\n\n", mxflow/*错误*/ + dinic());
	}
}

正确代码

#define _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include <cstring>
#include <queue>
#define int long long
using namespace std;

const int N = 370, M = 1e3 + 10, C = 3e2 + 10, INF = 0x3f3f3f3f;

int n, m, s, t, ss, tt;
struct node {
	int from, to, next, siz;
	node() {}
	node(int _from, int _to, int _next, int _siz) {
		from = _from, to = _to, next = _next, siz = _siz;
	}
}ed[(N * C + M) << 1];
int pre[N + M], rad[N + M];

void add(int x, int y, int siz) {
	ed[++pre[0]] = node(x, y, pre[x], siz);
	pre[x] = pre[0];
	ed[++pre[0]] = node(y, x, pre[y], 0);
	pre[y] = pre[0];
}

int tflow[N + M];

void clear() {
	memset(tflow, 0, sizeof(tflow));
	memset(pre, 0, sizeof(pre));
	pre[0] = 1;
}

int dep[N + M];
queue<int> q;

bool bfs() {
	memset(dep, 0, sizeof(dep));
	dep[s] = 1;
	q.push(s);
	while (!q.empty()) {
		int now = q.front();
		q.pop();
		rad[now] = pre[now];
		for (int i = pre[now]; i != 0; i = ed[i].next) {
			if (dep[ed[i].to] != 0 || ed[i].siz == 0) continue;
			dep[ed[i].to] = dep[now] + 1;
			q.push(ed[i].to);
		}
	}
	return dep[t] != 0;
}

int dfs(int now, int sum) {
	if (now == t) return sum;
	int val = sum;
	for (int i = rad[now]; i != 0; i = ed[i].next) {
		rad[now] = i;
		if (dep[ed[i].to] != dep[now] + 1 || ed[i].siz == 0) continue;
		int tmp = dfs(ed[i].to, min(val, ed[i].siz));
		ed[i].siz -= tmp;
		ed[i ^ 1].siz += tmp;
		val -= tmp;
		if (val == 0) break;
	}
	return sum - val;
}

int dinic() {
	int res = 0;
	while (bfs()) {
		res += dfs(s, INF);
	}
	return res;
}

signed main() {
	while (scanf("%lld%lld", &n, &m) != EOF) {
		clear();
		s = n + m + 1, t = n + m + 2, ss = n + m + 3, tt = n + m + 4;
		for (int i = 1; i <= m; i++) {
			int g;
			scanf("%lld", &g);
			add(n + i, t, INF);
			tflow[n + i] -= g;
			tflow[t] += g;
		}
		for (int i = 1; i <= n; i++) {
			int c, d;
			scanf("%lld%lld", &c, &d);
			add(s, i, d);
			for (int j = 1; j <= c; j++) {
				int _t, l, r;
				scanf("%lld%lld%lld", &_t, &l, &r);
				_t++;
				add(i, n + _t, r - l);
				tflow[i] -= l;
				tflow[n + _t] += l;
			}
		}
		int mxflow = 0;
		for (int i = 1; i <= t; i++)
			if (tflow[i] < 0) add(i, tt, -tflow[i]);
			else if (tflow[i] > 0) add(ss, i, tflow[i]), mxflow += tflow[i];
		add(t, s, INF);
		s = ss, t = tt;
		if (dinic() != mxflow) {
			printf("-1\n\n");
			continue;
		}
		int tmp = ed[pre[0]].siz;/*正确*/
		ed[pre[0]].siz = ed[pre[0] ^ 1].siz = 0;
		s = n + m + 1, t = n + m + 2;
		printf("%d\n\n", tmp + dinic());
	}
}