UVA 1221/HDU 2413/POJ 3343 Against Mammoths（二分+二分图匹配）

思路：

由于人类星球和外星球是一一对应的，自然想到二分图匹配，但是如果匹配，必须是某人类星球能打赢某外星球才连边。因为星球上的飞船数量随时间变化，所以先考虑把时间固定，然后就可以分类讨论求出在T时间内的最大匹配。由于在T时间内能胜利，那么大于T时间也一定能，二分即可。

细节参见代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 333*2;
int T,n,m;
struct Edge {
  int from, to, cap, flow;
};
bool operator < (const Edge& a, const Edge& b) {
  return a.from < b.from || (a.from == b.from && a.to < b.to);
}
struct Dinic {
  int n, m, s, t;        // 结点数， 边数（包括反向弧）， 源点编号， 汇点编号
  vector<Edge> edges;    // 边表， edges[e]和edges[e^1]互为反向弧
  vector<int> G[maxn];   // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
  bool vis[maxn];        // BFS使用
  int d[maxn];           // 从起点到i的距离
  int cur[maxn];         // 当前弧指针
void init(int n) {
    for(int i = 0; i < n; i++) G[i].clear();
    edges.clear();
}
void AddEdge(int from, int to, int cap) {
    edges.push_back((Edge){from, to, cap, 0});
    edges.push_back((Edge){to, from, 0, 0});
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
}
bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    vis[s] = 1;
    d[s] = 0;
    while(!Q.empty()) {
      int x = Q.front(); Q.pop();
      for(int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if(!vis[e.to] && e.cap > e.flow) {  //只考虑残量网络中的弧
          vis[e.to] = 1;
          d[e.to] = d[x] + 1;
          Q.push(e.to);
        }
      }
    }
    return vis[t];
}
int DFS(int x, int a) {
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int& i = cur[x]; i < G[x].size(); i++) {  //上次考虑的弧
      Edge& e = edges[G[x][i]];
      if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap-e.flow))) > 0) {
        e.flow += f;
        edges[G[x][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(a == 0) break;
      }
    }
    return flow;
}
int Maxflow(int s, int t) {
    this->s = s; this->t = t;
    int flow = 0;
    while(BFS()) {
      memset(cur, 0, sizeof(cur));
      flow += DFS(s, INF);
    }
    return flow;
  }
} g;
struct node {
    int init, add;
    node(int init=0, int add=0):init(init), add(add) {}
}a[maxn], b[maxn];
int t[maxn][maxn];
bool ok(int mid) {
    g.init(n+m+5);
    int src = 0, stc = n+m+1;
    for(int i = 1; i <= n; i++) g.AddEdge(src, i, 1);
    for(int i = 1; i <= m; i++) g.AddEdge(n+i, stc, 1);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            if(a[i].add <= b[j].add) {
                if(a[i].init >= (ll)b[j].init + t[i][j]*b[j].add) {
                    g.AddEdge(i, n+j, 1);
                }
            }
            else {
                ll t1 = mid - t[i][j];
                ll aa = (ll)a[i].init + t1*a[i].add;
                ll bb = (ll)b[j].init + t1*b[j].add;
                if(aa >= bb + (ll)t[i][j] * b[j].add) {
                    g.AddEdge(i, n+j, 1);
                }
            }
        }
    }
    int ans = g.Maxflow(src, stc);
    if(ans == m) return true;
    else return false;
}
int main() {
    while(~scanf("%d%d", &n, &m)) {
        if(n == 0 && m == 0) break;
        for(int i = 1; i <= n; i++) {
            scanf("%d%d", &a[i].init, &a[i].add);
        }
        for(int i = 1; i <= m; i++) {
            scanf("%d%d", &b[i].init, &b[i].add);
        }
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                scanf("%d", &t[i][j]);
            }
        }
        int l = 0, r = 1e9, mid;
        while(r > l) {
            mid = (l + r) >> 1;
            if(ok(mid)) r = mid;
            else l = mid + 1;
        }
        if(ok(l)) printf("%d\n", l);
        else printf("IMPOSSIBLE\n");
    }
    return 0;
}