I. Disks

合集 - 图论(79)

1.F. Chat Screenshots2024-02-192.P1656 炸铁路2024-02-223.P1137 旅行计划2024-02-244.P2835 刻录光盘2024-02-245.P1197 [JSOI2008] 星球大战2024-02-246.P3388 【模板】割点（割顶）2024-02-247.P8435 【模板】点双连通分量2024-02-258.P8436 【模板】边双连通分量2024-02-259.P2860 [USACO06JAN] Redundant Paths G2024-02-2510.P1653 [USACO04DEC] Cow Ski Area G2024-02-2611.P3047 [USACO12FEB] Nearby Cows G2024-03-0412.P1894 [USACO4.2] 完美的牛栏The Perfect Stall2024-03-0613.P1550 [USACO08OCT] Watering Hole G2024-03-0614.P2330 [SCOI2005] 繁忙的都市2024-03-0615.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0716.P1379 八数码难题2024-03-0717.P2746 [USACO5.3] 校园网Network of Schools2024-03-0818.P6121 [USACO16OPEN] Closing the Farm G2024-03-0819.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0820.P2055 [ZJOI2009] 假期的宿舍2024-03-0821.P5905 【模板】全源最短路（Johnson）2024-03-1122.F. Microcycle2024-03-1223.G. Path Prefixes2024-03-1424.G. Rudolf and Subway2024-03-1525.C. Ehab and Path-etic MEXs2024-03-1726.A. String Transformation 12024-03-1727.D. Secret Passwords2024-03-1928.F. Maximum White Subtree2024-03-1929.P3478 [POI2008] STA-Station2024-03-1930.P1347 排序2024-03-2131.P1960 郁闷的记者2024-03-2132.E1. Weights Division (easy version)2024-03-2233.P5007 DDOSvoid 的疑惑2024-03-2534.P2850 [USACO06DEC] Wormholes G2024-03-2635.P1265 公路修建2024-03-2836.P1354 房间最短路问题2024-03-2937.P2168 [NOI2015] 荷马史诗2024-03-3038.P8306 【模板】字典树2024-03-3039.P1481 魔族密码2024-03-3040.P3128 [USACO15DEC] Max Flow P2024-04-0241.P5536 【XR-3】核心城市2024-04-0242.P5836 [USACO19DEC] Milk Visits S2024-04-0243.P3384 【模板】重链剖分/树链剖分2024-04-0344.P5960 【模板】差分约束2024-04-0445.P7771 【模板】欧拉路径2024-04-0546.六度分离2024-04-0847.整数区间2024-04-0848.F. Alex's whims2024-04-1249.J. 上学2024-05-1450.Game on Tree2024-05-1451.E. We Need More Bosses2024-05-1552.B. Omkar and Heavenly Tree2024-05-1653.B. Mahmoud and Ehab and the bipartiteness2024-05-1654.P1668 [USACO04DEC] Cleaning Shifts S2024-05-1755.P6154 游走2024-05-2056.P8655 [蓝桥杯 2017 国 B] 发现环2024-05-2457.P10298 [CCC 2024 S4] Painting Roads2024-05-2458.P9650 [SNCPC2019] Escape Plan2024-05-2759.P9327 [CCC 2023 S4] Minimum Cost Roads2024-05-2960.P9026 [CCC2021 S4] Daily Commute2024-05-2961.P8724 [蓝桥杯 2020 省 AB3] 限高杆2024-06-0362.P4878 [USACO05DEC] Layout G2024-06-0463.P5663 [CSP-J2019] 加工零件2024-06-0464.P2731 [USACO3.3] 骑马修栅栏 Riding the Fences2024-06-08

65.I. Disks2024-06-17

66.P1351 [NOIP2014 提高组] 联合权值2024-06-2667.B. Time Travel2024-06-2968.F. Minimum Maximum Distance2024-07-1469.A. Book2024-07-1970.P1407 [国家集训队] 稳定婚姻2024-07-2071.P1991 无线通讯网2024-07-2372.P4047 [JSOI2010] 部落划分2024-07-2373.P3275 [SCOI2011] 糖果2024-07-2374.P1989 无向图三元环计数2024-07-2775.P1967 [NOIP2013 提高组] 货车运输2024-07-3176.D. Vitaly and Cycle2024-08-0177.P10838 『FLA - I』庭中有奇树2024-08-0778.P9751 [CSP-J 2023] 旅游巴士2024-08-1179.D. Colored Portals2024-08-21

收起

原题链接

题解

对于一组相切的圆来说，其中一个圆变大，其相邻的圆变小，然后相邻的相邻的圆变大...而要让总半径和变小，一定得是总的变小的圆更多

实施

先判断一组圆能不能发生变化，然后再累积变大和变小的圆个数

code

#include<bits/stdc++.h>
using namespace std;

#define ll long long
const ll inf=2e14;

struct node
{
    ll x,y,r;
}c[1005];

vector<ll> G[1005];

ll vis[1005]={0};

inline void read(ll &x) {
	x = 0;
	ll flag = 1;
	char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-')flag = -1;
        c = getchar();
    }
	while(c >= '0' && c <= '9') {
		x = (x << 3) + (x << 1) + (c ^ 48);
		c = getchar();
	}
	x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
    	putchar('-');
		x = -x;
	}
    if(x > 9)
		write(x / 10);
    putchar(x % 10 + '0');
}

ll res=0;
void dfs(ll now, ll fa, ll op)
{
    res+=op;
    vis[now]=op;
    for(auto next:G[now])
    {
        if(next==fa) continue;
        if(!vis[next])
        {
            dfs(next,now,-op);
        }
        else if(vis[next]==vis[now]) res+=inf;
    }

}

int main()
{
    ll n;
    read(n);
    for(ll i=1; i<=n; i++)
    {
        read(c[i].x);
        read(c[i].y);
        read(c[i].r);
    }

    for(ll i=1; i<=n; i++)
    {
        for(ll j=1; j<i; j++)
        {
            ll x1=c[i].x, x2=c[j].x, y1=c[i].y, y2=c[j].y, r1=c[i].r, r2=c[j].r;
            if((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)==(r1+r2)*(r1+r2))
            {
                G[i].push_back(j);
                G[j].push_back(i);
            }
        }
    }

    for(ll i=1; i<=n; i++)
    {
        if(!vis[i])
        {
            res=0;
            dfs(i,i,1);
            if(res&&res<=n)
            {
                puts("YES");
                return 0;
            }
        }
    }

    puts("NO");
    return 0;
}