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);} //head int 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; }