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76 lines
2.2 KiB
76 lines
2.2 KiB
#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 100010, M = 300010;
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//与AcWing 1169. 糖果 这道题一模一样，连测试用例都一样
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stack<int> q; //有时候换成栈判断环很快就能
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LL dist[N];
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bool st[N];
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int cnt[N];
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int n, m; //表示点数和边数
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//邻接表
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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bool spfa() { //求最长路，所以判断正环
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memset(dist, -0x3f, sizeof dist); //初始化为-0x3f
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//差分约束从超级源点出发
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dist[0] = 0;
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q.push(0);
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st[0] = true;
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while (q.size()) {
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int u = q.top();
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q.pop();
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st[u] = false;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (dist[j] < dist[u] + w[i]) { //求最长路
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dist[j] = dist[u] + w[i];
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cnt[j] = cnt[u] + 1;
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//注意多加了超级源点到各各节点的边
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if (cnt[j] >= n + 1) return false;
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if (!st[j]) {
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q.push(j);
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st[j] = true;
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}
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}
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}
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}
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return true;
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}
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int main() {
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memset(h, -1, sizeof h);
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scanf("%d %d", &n, &m);
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for (int i = 0; i < m; i++) {
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int op, a, b; // op为选择
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scanf("%d %d %d", &op, &a, &b);
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if (op == 1) /** a == b => (a >= b , b >= a) */
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add(a, b, 0), add(b, a, 0);
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else if (op == 2) /** b >= a + 1 */
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add(a, b, 1);
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else if (op == 3) /** a >= b */
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add(b, a, 0);
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else if (op == 4) /** a >= b + 1 */
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add(b, a, 1);
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else /** b >= a */
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add(a, b, 0);
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}
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/** xi >= x0 + 1 （每个小朋友都要至少一个糖果）*/
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//将所有节点与超级源点x0相连
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for (int i = 1; i <= n; ++i) add(0, i, 1);
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if (!spfa())
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puts("-1");
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else {
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LL res = 0;
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for (int i = 1; i <= n; ++i) res += dist[i];
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printf("%lld\n", res);
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}
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return 0;
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} |