【二分匹配】 ACdream 1403 Graph Game

先做二分匹配，然后遍历每个点，如果删掉这个点求出的最大匹配和原先的最大匹配相等，那么这个点就是必败点，否则就是必胜点。。。

#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 1005#define maxm 100005#define eps 1e-10#define mod 10000007#define INF 1e9#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid  #define rson o<<1 | 1, mid+1, R  typedef long long LL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}//headint h[maxn], next[maxm], v[maxm];int linked[maxn], vis[maxn], res[maxn];int n1, n2, m, cnt, p;void addedges(int a, int b){next[cnt] = h[a], h[a] = cnt, v[cnt] = b, cnt++;}void init(void){cnt = p = 0;memset(h, -1, sizeof h);memset(linked, -1, sizeof linked);memset(res, -1, sizeof res);}bool find(int u){for(int e = h[u]; ~e; e = next[e]) if(!vis[v[e]]) {vis[v[e]] = 1;if(linked[v[e]] == -1 || find(linked[v[e]])) {if(v[e] == p) continue;linked[v[e]] = u;res[u] = v[e];return true;}}return false;}void read(void){int a, b;while(m--) {scanf("%d%d", &a, &b);addedges(a, b+n1);addedges(b+n1, a);}}void work(void){for(int i = n1+1; i <= n1+n2; i++) {memset(vis, 0, sizeof vis);find(i);}for(int i = 1; i <= n1; i++) {if(linked[i] == -1) printf("P");else {int t = linked[i];linked[i] = res[t] = -1;p = i;memset(vis, 0, sizeof vis);if(find(t)) printf("P");else printf("N"), res[t] = i, linked[i] = t;}}printf("\n");memset(linked, -1, sizeof linked);memset(res, -1, sizeof res);for(int i = 1; i <= n1; i++) {memset(vis, 0, sizeof vis);find(i);}for(int i = n1+1; i <= n1+n2; i++) {if(linked[i] == -1) printf("P");else {int t = linked[i];linked[i] = res[t] = -1;p = i;memset(vis, 0, sizeof vis);if(find(t)) printf("P");else printf("N"), res[t] = i, linked[i] = t;}}printf("\n");}int main(void){while(scanf("%d%d%d", &n1, &n2, &m)!=EOF) {init();read();work();}return 0;}