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;
// & 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;
}