P1640 [SCOI2010]连续攻击游戏 二分图构造

时间:2024-09-28 13:34:32

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> /* ⊂_ヽ
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'ノ )  Lノ */ 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<<;
const ll mos = 0x7FFFFFFF; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e9+;
const double esp = 1e-;
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=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
}
struct FastIO {
static const int S = 4e6;
int wpos;
char wbuf[S];
FastIO() : wpos() {}
inline int xchar() {
static char buf[S];
static int len = , pos = ;
if (pos == len)
pos = , len = fread(buf, , S, stdin);
if (pos == len) exit();
return buf[pos++];
}
inline int xuint() {
int c = xchar(), x = ;
while (c <= ) c = xchar();
for (; '' <= c && c <= ''; c = xchar()) x = x * + c - '';
return x;
}
inline int xint()
{
int s = , c = xchar(), x = ;
while (c <= ) c = xchar();
if (c == '-') s = -, c = xchar();
for (; '' <= c && c <= ''; c = xchar()) x = x * + c - '';
return x * s;
}
inline void xstring(char *s)
{
int c = xchar();
while (c <= ) c = xchar();
for (; c > ; c = xchar()) * s++ = c;
*s = ;
}
inline void wchar(int x)
{
if (wpos == S) fwrite(wbuf, , S, stdout), wpos = ;
wbuf[wpos++] = x;
}
inline void wint(int x)
{
if (x < ) wchar('-'), x = -x;
char s[];
int n = ;
while (x || !n) s[n++] = '' + x % , x /= ;
while (n--) wchar(s[n]);
wchar('\n');
}
inline void wstring(const char *s)
{
while (*s) wchar(*s++);
}
~FastIO()
{
if (wpos) fwrite(wbuf, , wpos, stdout), wpos = ;
}
} 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 = ;
struct E{
int v,nxt;
}edge[];
int head[maxn],gtot;
void addedge(int u,int v){
edge[gtot].v = v;
edge[gtot].nxt = head[u];
head[u] = gtot++;
}
int used[],pt[]; 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]== || hungry(pt[v],col)){
pt[v] = u;
return true;
}
}
}
return false;
}
int main(){
int n; scanf("%d", &n);
memset(head, -, sizeof(head));
for(int i=; i<=n; i++){
int x,y;
scanf("%d%d", &x, &y);
addedge(x, i);
addedge(y, i);
}
int ans = ,col = ;
for(int i=; i<=; i++){
++col;
if(hungry(i,col)) ans = i;
else break;
}
printf("%d\n", ans);
return ;
}