UVA - 11478 Halum （最短路應用+二分）

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Description

	Problem H	Halum	Time Limit : 3 seconds
	You are given a directed graph G(V,E) with a set of vertices and edges. Each edge (i,j) that connects some vertex i to vertex j has an integer cost associated with that edge.   Define the operation Halum(v, d) to operate on a vertex v using an integer d as follows: subtract d from the cost of all edges that enter v and add d to the cost of every edge that leaves v. As an example of that operation, consider graph G that has three vertices named (1, 2, 3) and two edges. Edge (1, 2) has cost -1, and edge (2,3) has cost 1. The operation Halum(2,-3) operates on edges entering and leaving vertex 2.  Thus, edge (1, 2) gets cost -1-(-3)=2 and the edge (2, 3) gets cost 1 + (-3) = -2. Your goal is to apply the Halum function to a graph, potentially repeatedly, until every edge in the graph has at least a certain cost that is greater than zero. You have to maximize this cost.
	Input
	Two space-separated integers per case: V(V≤500) andE(E≤2700). E lines follow. Each line represents a directed edge using three space-separated integers (u, v, d). Absolute value of cost can be at most 10000.
	Output
	If the problem is solvable, then print the maximum possible value. If there is no such solution print “No Solution”. If the value can be arbitrary large print “Infinite”
	Sample Input	Sample Output
	2 1 1 2 10 2 1 1 2 -10 3 3 1 2 4 2 3 2 3 1 5 4 5 2 3 4 4 2 5 3 4 2 3 1 0 1 2 -1	Infinite Infinite 3 1

題意：給定一個有向圖，每條邊都有一個權值，每次你可選擇一個結點v和一個整數d，把所有以v為終點的邊的權值減少d，把所有以v為終點的邊的權值增加d，最後要讓所有邊的最小值非負且盡量大。

思路：最小值最大，很顯然想到二分答案，可以令sum(u)表示為作用在節點u之上的所有d之和，這樣題目的目標就是確定所有的sum(u)了； 對於邊a->b，不難發現操作後的權值是:w(a, b)+sum(a)-sum(b)>=x

那麼就可以得到個不等式：sum(b)-sum(a) <= w(a, b)-x，而w(a, b)-x我們是已知的，那麼就能得到相當於最短路的

不等式 d[v]<=d[u]+w(u, v)，那麼我們就可以構建一個新圖，那麼我們在做Bellman-Ford的時候，如果發現負權環的話，那麼就相當於我們無法得到一個類似最短路的不等式，所有無解

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <queue>#include <vector>using namespace std;const int maxn = 505;const int inf = 0x3f3f3f3f;struct Edge {int from, to, dist;};struct BellmanFord {int n, m;vector<Edge> edges;vector<int> G[maxn];int inq[maxn];int d[maxn];int p[maxn];int cnt[maxn];void init(int n) {this->n = n;for (int i = 0; i < n; i++)G[i].clear();edges.clear();}void AddEdge(int from, int to, int dist) {edges.push_back((Edge){from, to, dist});m = edges.size();G[from].push_back(m-1);}bool negativeCycle(int tmp) {queue<int> Q;memset(inq, 0, sizeof(inq));memset(cnt, 0, sizeof(cnt));for (int i = 0; i < n; i++) {d[i] = 0;inq[0] = 1;Q.push(i);}while (!Q.empty()) {int u = Q.front();Q.pop();inq[u] = 0;for (int i = 0; i < G[u].size(); i++) {Edge &e = edges[G[u][i]];if (d[e.to] > d[u] + e.dist - tmp) {d[e.to] = d[u] + e.dist - tmp;p[e.to] = G[u][i];if (!inq[e.to]) {Q.push(e.to);inq[e.to] = 1;if (++cnt[e.to] > n)return 1;}}}}return 0;}};BellmanFord solve;int n, m;int main() {int u, v, w;while (scanf("%d%d", &n, &m) != EOF) {solve.init(n);int l = 1, r = 0, mid;for (int i = 0; i < m; i++) {scanf("%d%d%d", &u, &v, &w);u--, v--;solve.AddEdge(u, v, w);r = max(r, w);}if (solve.negativeCycle(1))printf("No Solution\n");else if (!solve.negativeCycle(r+1))printf("Infinite\n");else {while (l < r) {mid = l + (r-l+1)/2;if (solve.negativeCycle(mid))r = mid-1;else l = mid;}printf("%d\n", l);}}return 0;}