P1640 [SCOI2010]连续攻击游戏 二分图构造
题意
lxhgww最近迷上了一款游戏,在游戏里,他拥有很多的装备,每种装备都有2个属性,这些属性的值用[1,10000]之间的数表示。当他使用某种装备时,他只能使用该装备的某一个属性。并且每种装备最多只能使用一次。游戏进行到最后,lxhgww遇到了终极boss,这个终极boss很奇怪,攻击他的装备所使用的属性值必须从1开始连续递增地攻击,才能对boss产生伤害。也就是说一开始的时候,lxhgww只能使用某个属性值为1的装备攻击boss,然后只能使用某个属性值为2的装备攻击boss,然后只能使用某个属性值为3的装备攻击boss……以此类推。现在lxhgww想知道他最多能连续攻击boss多少次?
武器的个数<=1000000
思路
这个构图我觉得是比较巧妙的。单单拆点按不同属性分两边不太好想。
这道题合理的二分图中,左边1~10000表示攻击的序列,右边1~n表示武器。从左边向右边对应武器连两条有向边,跑二分图匹配,就很巧妙的使得这两条边不会同时成立。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> using namespace std; #define lson (l, mid, rt << 1) #define rson (mid + 1, r, rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll, ll> pll; typedef pair<int, int> pii; typedef pair<int, pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define boost \ ios::sync_with_stdio(false); \ cin.tie(0) #define rep(a, b, c) for (int a = (b); a <= (c); ++a) #define max3(a, b, c) max(max(a, b), c); #define min3(a, b, c) min(min(a, b), c); const ll oo = 1ll << 17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9 + 7; const double esp = 1e-8; const double PI = acos(-1.0); const double PHI = 0.61803399; //黄金分割点 const double tPHI = 0.38196601; template <typename T> inline T read(T &x) { x = 0; int f = 0; char ch = getchar(); while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x = f ? -x : x; } struct FastIO { static const int S = 4e6; int wpos; char wbuf[S]; FastIO() : wpos(0) { } inline int xchar() { static char buf[S]; static int len = 0, pos = 0; if (pos == len) pos = 0, len = fread(buf, 1, S, stdin); if (pos == len) exit(0); return buf[pos++]; } inline int xuint() { int c = xchar(), x = 0; while (c <= 32) c = xchar(); for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0'; return x; } inline int xint() { int s = 1, c = xchar(), x = 0; while (c <= 32) c = xchar(); if (c == '-') s = -1, c = xchar(); for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0'; return x * s; } inline void xstring(char *s) { int c = xchar(); while (c <= 32) c = xchar(); for (; c > 32; c = xchar()) *s++ = c; *s = 0; } inline void wchar(int x) { if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0; wbuf[wpos++] = x; } inline void wint(int x) { if (x < 0) wchar('-'), x = -x; char s[24]; int n = 0; while (x || !n) s[n++] = '0' + x % 10, x /= 10; while (n--) wchar(s[n]); wchar('\n'); } inline void wstring(const char *s) { while (*s) wchar(*s++); } ~FastIO() { if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0; } } io; inline void cmax(int &x, int y) { if (x < y) x = y; } inline void cmax(ll &x, ll y) { if (x < y) x = y; } inline void cmin(int &x, int y) { if (x > y) x = y; } inline void cmin(ll &x, ll y) { if (x > y) x = y; } /*-----------------------showtime----------------------*/ const int maxn = 10009; struct E { int v, nxt; } edge[2000009]; int head[maxn], gtot; void addedge(int u, int v) { edge[gtot].v = v; edge[gtot].nxt = head[u]; head[u] = gtot++; } int used[1000009], pt[1000009]; bool hungry(int u, int col) { for (int i = head[u]; ~i; i = edge[i].nxt) { int v = edge[i].v; if (used[v] < col) { used[v] = col; if (pt[v] == 0 || hungry(pt[v], col)) { pt[v] = u; return true; } } } return false; } int main() { int n; scanf("%d", &n); memset(head, -1, sizeof(head)); for (int i = 1; i <= n; i++) { int x, y; scanf("%d%d", &x, &y); addedge(x, i); addedge(y, i); } int ans = 0, col = 0; for (int i = 1; i <= 10000; i++) { ++col; if (hungry(i, col)) ans = i; else break; } printf("%d\n", ans); return 0; }
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