CSU-ACM2017暑假集训比赛8

E - Escape

 2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.

Input

    More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.    The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..    0 <= ai <= 100000

Output

    Determine whether all people can live up to these stars    If you can output YES, otherwise output NO.

Sample Input

1 1112 21 01 01 1

Sample Output

YESNO

进行状态压缩后跑最大流即可

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <queue>using namespace std;const int maxn = 3004, INF = 0x3f3f3f3f;struct Edge {    int from, to, cap, flow;    Edge(){}    Edge(int ff, int t, int c, int f):from(ff), to(t), cap(c), flow(f){}};int N, M;vector<Edge> edges;vector<int> G[maxn];bool vis[maxn];int d[maxn], cur[maxn];struct Dinic {    int n, m, s, t;    void init() {        for(int i = 0; i <= maxn; i++)            G[i].clear();        edges.clear();        memset(vis, false, sizeof(vis));        memset(d, 0, sizeof(d));        memset(cur, 0, sizeof(cur));    }    void makeEdge(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);        d[s] = 0;        vis[s] = true;        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) {                    vis[e.to] = true;                    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++) {            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;    }};int main() {#ifdef TEST    freopen("test.txt", "r", stdin);#endif // TEST    char res[][4] {"NO", "YES"};    while(cin >> N >> M) {        Dinic Gf;        Gf.init();        int source = 0, terminal = (1<<M)+M+1;        int status[1080], temp;        memset(status, 0, sizeof(status));        for(int i = 0; i < N; i++){            int S = 0;            for(int j = 0; j < M; j++){                scanf("%d", &temp);                S |= (temp<<(M-1-j));            }            status[S]++;        }        for(int i = 1; i <= (1<<M); i++){            if(status[i]){                Gf.makeEdge(source, i, status[i]);                int x = i, dig = M;                while(x){                    if(x & 1)                        Gf.makeEdge(i, (1<<M)+dig, INF);                    x >>= 1;                    dig--;                }//                for(int j = 0; j < M; j++)//                    if(i & (1<<j))//                        Gf.makeEdge(i, (1<<M)+j+1, INF);            }        }        for(int i = 1; i <= M; i++){            int capability;            scanf("%d", &capability);            Gf.makeEdge((1<<M)+i, terminal, capability);        }        int mf = Gf.maxFlow(source, terminal);        printf("%s\n", res[mf == N]);    }    return 0;}