Hdu 3605 Escape

大意不再赘述。

思路：开始直接建图TLE啦，后来去网上查查发现要用状态压缩DP，由于第一次接触状态DP，想了很久的时间。还过几天，我要把整个12月的时间用来弄DP啦。

说下我的理解，首先最多有10个星球，对于每个人来说，每个星球Mi的选取状态为选或不选，选用1表示，不选用0表示，所以最多有2^10个状态，化为10进制的值的范围是0~1023.

即，我们可以把100000点压缩成1024个点，我们把相同状态的人的个数（表现为sum的值相同）放在一个节点的点权中，极限情况下，边数可以从原来的2000000条边压缩为1024*10*2，这样时间效率自然就上去了。

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <queue>#include <stack>using namespace std;const int MAXN = 20000;const int MAXM = 300010;const int INF = 0x3f3f3f3f;struct Edge{int v, f;int next;}edge[MAXM];int n, m;int cnt;int s, t;int first[MAXN], level[MAXN];int q[MAXN];int val[MAXN];int dp[MAXN];void init(){cnt = 0;memset(first, -1, sizeof(first));memset(dp, 0, sizeof(dp));}void read(int u, int v, int f){edge[cnt].v = v, edge[cnt].f = f;edge[cnt].next = first[u], first[u] = cnt++;}void read_graph(int u, int v, int f){read(u, v, f);read(v, u, 0);}int bfs(int s, int t){memset(level, 0, sizeof(level));level[s] = 1;int front = 0, rear = 1;q[front] = s;while(front < rear){int x = q[front++];if(x == t) return 1;for(int e = first[x]; e != -1; e = edge[e].next){int v = edge[e].v, f = edge[e].f;if(!level[v] && f){level[v] = level[x] + 1;q[rear++] = v;}}}return 0;}int dfs(int u, int maxf, int t){if(u == t) return maxf;int ret = 0;for(int e = first[u]; e != -1; e = edge[e].next){int v = edge[e].v, f = edge[e].f;if(level[v] == level[u] + 1 && f){int Min = min(maxf-ret, f);f = dfs(v, Min, t);edge[e].f -= f;edge[e^1].f += f;ret += f;if(ret == maxf) return ret;}}return ret;}int Dinic(int s, int t){int ans = 0;while(bfs(s, t)) ans += dfs(s, INF, t);return ans;}void build(){init();s = 0, t = (1<<m)+m+1;for(int i = 0; i < n; i++){int sum = 0;for(int j = 0; j < m; j++){int flag;scanf("%d", &flag);if(flag) sum += (1<<j);}dp[sum]++;}for(int i = 0; i < m; i++){int f;scanf("%d", &f);read_graph((1<<m)+i, t, f); //星球连向汇点,容量为限制度 }for(int i = 0; i < (1<<m); i++) if(dp[i]){read_graph(s, i, dp[i]); //源点连向人，容量为dp[i] for(int j = 0; j < m; j++){if(i & (1<<j)){read_graph(i, (1<<m)+j, dp[i]); //人连星球，容量为dp[i] }}}}void solve(){build();int ans = Dinic(s, t);if(ans >= n) printf("YES\n");else printf("NO\n");}int main(){while(~scanf("%d%d", &n, &m)){solve();}return 0;}