BZOJ1050 [HAOI2006]旅行

其实这道题根本不用最短路算法...

我们可以就把边从小到大排序,那么只需要枚举大小两个端点,把中间的边都加进去判断联通性即可.

判断联通性显然用的是并查集.

#include <cstdio>
#include <algorithm>
struct DJSET{
	struct{
		int fa,rk;
	} p[505];
	int find(int k){ return p[k].fa?p[k].fa=find(p[k].fa):k; }
	inline void merge(int a,int b){
		a=find(a),b=find(b);
		if(a==b) return;
		if(p[a].rk<p[b].rk) std::swap(a,b);
		p[b].fa=a;
		if(p[a].rk==p[b].rk) ++p[a].rk;
	}
} set[5005];
struct eg{
	int f,t,l;
	inline void read(){
		scanf("%d%d%d",&f,&t,&l);
	}
} Es[5005];
int gcd(int a,int b){
	if(a) return gcd(b%a,a);
	return b;
}
bool cmp(const eg& a,const eg& b){
	return a.l<b.l;
}
int m,n,i,j,s,t;
int FA,FB,FC,FA1,FB1,FC1;
inline void reduce(){
	FC=gcd(FA,FB);
	FA/=FC,FB/=FC;
}
int main(){
	scanf("%d%d",&n,&m);
	for(i=0;i<m;++i) Es[i].read();
	std::sort(Es,Es+m,cmp);
	scanf("%d%d",&s,&t);
	FA1=1000;
	for(i=0;i<m;++i){
		for(j=i;j<m;++j){
			set[i].merge(Es[j].f,Es[j].t);
			if(set[i].find(s)==set[i].find(t)){
				FA=Es[j].l,FB=Es[i].l;
				reduce();
				if(FA*FB1<FB*FA1){
					FA1=FA;
					FB1=FB;
				}
				break;
			}
		}
	}
	if(FB1==0) printf("IMPOSSIBLE"); else{
		if(FB1==1) printf("%d\n",FA1); else printf("%d/%d\n",FA1,FB1);
	}
	return 0;
}

其实有个小优化:当所有边都加入而不联通时直接退出IMPOSSIBLE.

然后哎唷我擦快了不少...

#include <cstdio>
#include <algorithm>
struct DJSET{
	struct{
		int fa,rk;
	} p[505];
	int find(int k){ return p[k].fa?p[k].fa=find(p[k].fa):k; }
	inline void merge(int a,int b){
		a=find(a),b=find(b);
		if(a==b) return;
		if(p[a].rk<p[b].rk) std::swap(a,b);
		p[b].fa=a;
		if(p[a].rk==p[b].rk) ++p[a].rk;
	}
} set[5005];
struct eg{
	int f,t,l;
	inline void read(){
		scanf("%d%d%d",&f,&t,&l);
	}
} Es[5005];
int gcd(int a,int b){
	if(a) return gcd(b%a,a);
	return b;
}
bool cmp(const eg& a,const eg& b){
	return a.l<b.l;
}
int m,n,i,j,s,t;
int FA,FB,FC,FA1,FB1,FC1;
inline void reduce(){
	FC=gcd(FA,FB);
	FA/=FC,FB/=FC;
}
int main(){
	scanf("%d%d",&n,&m);
	for(i=0;i<m;++i) Es[i].read();
	std::sort(Es,Es+m,cmp);
	scanf("%d%d",&s,&t);
	FA1=1000;
	for(i=0;i<m;++i){
		for(j=i;j<m;++j){
			set[i].merge(Es[j].f,Es[j].t);
			if(set[i].find(s)==set[i].find(t)){
				FA=Es[j].l,FB=Es[i].l;
				reduce();
				if(FA*FB1<FB*FA1){
					FA1=FA;
					FB1=FB;
				}
				break;
			}
		}
		if(FB1==0){
			printf("IMPOSSIBLE\n");
			return 0;
		}
	}
	if(FB1==1) printf("%d\n",FA1); else printf("%d/%d\n",FA1,FB1);
	return 0;
}

由于空间够大我还偷了个懒没有每次memset一遍...然后就出现了奇异的景象:

Submit	1737
User(Submit)	735
User(Solved)	702
AC	871
WA	460
TLE	196
MLE	3
OLE	6
RE	143
CE	58
AC[50%]   WA[26%]   TLE[11%]   MLE[0%]   OLE[0%]   RE[8%]   CE[3%]

No.	RunID	User	Memory	Time	Language	Code_Length	Submit_Time
241	546203	absi2011	1468 KB	1076 MS	C++	2608 B	2014-02-14 16:51:06
242	695809	zhengly123	1824 KB	1076 MS	C++	1730 B	2014-07-15 14:58:06
243	473121	sth1997	864 KB	1088 MS	C++	1321 B	2013-08-29 10:59:59
244	967344(2)	tmzbot	20616 KB	1116 MS	C++	1311 B	2015-05-08 14:26:09
245	312288	Maigo	1352 KB	1144 MS	C++	1060 B	2012-10-06 16:16:30
246	234103	Oct_Gap201	372 KB	1152 MS	Pascal	4710 B	2012-03-27 16:40:05
247	786906	Nano_Ape	1480 KB	1152 MS	C++	1705 B	2014-11-26 14:35:22
248	6697(2)	Sachs	700 KB	1153 MS	Pascal	2155 B	2009-09-09 18:39:30
249	221303	TSZWL	368 KB	1156 MS	Pascal	4662 B	2012-03-10 16:07:32
250	244488(2)	bill125	368 KB	1156 MS	Pascal	4662 B	2012-04-16 23:08:22
251	214563(2)	sbsbsbsb	368 KB	1160 MS	Pascal	4670 B	2012-02-28 21:34:20
252	197945	This_poet	1396 KB	1168 MS	C++	1564 B	2012-02-01 17:13:54
253	67605	nathenqian	268 KB	1183 MS	C++	1737 B	2010-12-11 14:49:13
254	68313	cfttcc	860 KB	1183 MS	Pascal	1751 B	2010-12-14 13:33:07
255	75681	crfish	600 KB	1184 MS	Pascal	1662 B	2011-02-25 23:25:16
256	29182	wu3412790	620 KB	1184 MS	Pascal	1381 B	2010-06-11 17:00:14
257	661385	RealCS	1396 KB	1188 MS	C++	1560 B	2014-05-30 15:37:42
258	12275	thegreysky	320 KB	1199 MS	C++	1523 B	2010-03-09 22:25:46
259	6500	zr_1234	684 KB	1201 MS	Pascal	1957 B	2009-09-01 18:26:15
260	460098	Ruchiose	1332 KB	1212 MS	C++	2924 B	2013-08-02 17:20:09

[TOP]  [STATUS]  [PREV]  [NEXT]

2W多KB的那个SB就是我.