UVA 11478 - Halum（差分約束+最短路)

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UVA 11478 - Halum

題目連結

題意：給定一個有向圖，每次操作可以選擇一個結點，把以這個點為起點的邊權值+d，以這個邊為終點的-d，問經過操作後，能得到的邊權最小的最大值是多少，並且要判但是否無窮大或無解

思路：轉化為差分約束，設一條邊，他增加的權值為sum(u)減少了sum(v)，那麼二分答案x，得到一個不等式sum(u) - sum(v) + w(u, v) >= x，變形後得到sum(v) - sum(u) <= w(u, v) - x，這樣就轉化為了差分約束，直接bellmenford判負環即可

代碼：

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#include <queue>using namespace std;typedef int Type;const int MAXNODE = 505;const int MAXEDGE = 2777;struct Edge {int u, v;Type dist;Edge() {}Edge(int u, int v, Type dist) {this->u = u;this->v = v;this->dist = dist;}};struct BellmanFrod {int n, m;Edge edges[MAXEDGE];int first[MAXNODE];int next[MAXEDGE];bool inq[MAXNODE];Type d[MAXNODE];int p[MAXNODE];int cnt[MAXNODE];void init(int n) {this->n = n;memset(first, -1, sizeof(first));m = 0;}void add_Edge(int u, int v, Type dist) {edges[m] = Edge(u, v, dist);next[m] = first[u];first[u] = m++;}bool negativeCycle() {queue<int> Q;memset(inq, 0, sizeof(inq));memset(cnt, 0, sizeof(cnt));for (int i = 0; i < n; i++) {d[i] = 0; inq[i] = true; Q.push(i);}while (!Q.empty()) {int u = Q.front();Q.pop();inq[u] = false;for (int i = first[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (d[e.v] > d[u] + e.dist) {d[e.v] = d[u] + e.dist;p[e.v] = i;if (!inq[e.v]) {Q.push(e.v);inq[e.v] = true;if (++cnt[e.v] > n) return true;}}}}return false;}} gao;int n, m;bool judge(int x) {for (int i = 0; i < gao.m; i++)gao.edges[i].dist -= x;bool tmp = gao.negativeCycle();for (int i = 0; i < gao.m; i++)gao.edges[i].dist += x;return tmp;}int main() {while (~scanf("%d%d", &n, &m)) {gao.init(n);int u, v, dist;int l = 1, r = 0;while (m--) {scanf("%d%d%d", &u, &v, &dist);u--, v--; r = max(r, dist);gao.add_Edge(u, v, dist);}r++;if (judge(l)) printf("No Solution\n");else if (!judge(r)) printf("Infinite\n");else {while (l < r) {int mid = (l + r) / 2;if (judge(mid)) r = mid;else l = mid + 1;}printf("%d\n", l - 1);}}return 0;}

UVA 11478 - Halum（差分約束+最短路)