UVa 11478 Halum (差分约束)

题意：给定一个有向图，每条边都有一个权值，每次你可以选择一个结点v和整数d，把所有以v为终点的边权值减少d,把所有以v为起点的边权值增加d，最后要让所有的边权值非负且最大。

析：首先二分答案，很容易想到，令sum(u) 表示作用在 u 结点的所有d的和，然后对于一条 u 到 v 的边，要满足大于 ans，也就是sum(v) - sum(u) < w(u, v) - ans，然后可以形成很多个不等式组，就是可以用差分约束来做了，每次只要判断是不是有负环就好了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 500 + 10;
const int mod = 1000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r > 0 && r <= n && c > 0 && c <= m;
}

struct Edge{
  int from, to, dist;
};

struct BellmanFord{
  int n, m;
  vector<Edge> edges;
  vector<int> G[maxn];
  int inq[maxn];
  int cnt[maxn];
  int d[maxn];

  void init(int n){
    this-> n = n;
    for(int i = 0; i < n; ++i)  G[i].cl;
    edges.cl;
  }

  void addEdge(int u, int v, int d){
    edges.pb((Edge){u, v, d});
    m = edges.sz;
    G[u].pb(m-1);
  }

  bool bfs(){
    queue<int> q;
    ms(inq, 0);  ms(cnt, 0);
    ms(d, 0);  inq[0] = true;
    for(int i = 0; i < n; ++i)  q.push(i);

    while(!q.empty()){
      int u = q.front();  q.pop();
      inq[u] = 0;
      for(int i = 0; i < G[u].sz; ++i){
        Edge &e = edges[G[u][i]];
        if(d[e.to] > d[u] + e.dist){
          d[e.to] = d[u] + e.dist;
          if(!inq[e.to]){
            q.push(e.to);
            inq[e.to] = 1;
            if(++cnt[e.to] > n)  return true;
          }
        }
      }
    }
    return false;
  }

  bool solve(int x){
    for(int i = 0; i < edges.sz; ++i)
      edges[i].dist -= x;
    bool ans = bfs();
    for(int i = 0; i < edges.sz; ++i)
      edges[i].dist += x;
    return ans;
  }
};

BellmanFord bell;

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    bell.init(n);
    int l = 1, r = 0;
    for(int i = 0; i < m; ++i){
      int u, v, c;
      scanf("%d %d %d", &u, &v, &c);
      --u, --v;
      bell.addEdge(u, v, c);
      r = max(r, c);
    }
    if(bell.solve(l))  puts("No Solution");
    else if(!bell.solve(r+1))  puts("Infinite");
    else{
      while(l <= r){
        int m = l + r >> 1;
        if(bell.solve(m))  r = m - 1;
        else l = m + 1;
      }
      printf("%d\n", l-1);
    }
  }
  return 0;
}