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

题目链接：点击打开链接

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;// & 0x7FFFFFFFconst 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;}