hihocoder 1089：最短路（Floyd）

hihocoder 1089

题解：最短路Floyd裸题

代码：

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
int const inf = 0x7f7f7f7f;
int const N = 100 + 10;
int n,m,s,t,d[N];
int mp[N][N];
set<pii>st;
void Floyd(){
	for(int k=1;k<=n;k++){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++)
				if(mp[i][j] > mp[i][k] + mp[k][j] && mp[i][k] != inf && mp[k][j] != inf)
					mp[i][j] = mp[i][k] + mp[k][j];
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(i == j)	mp[i][j] = 0;
			else mp[i][j] = inf;
		}
	}
	st.clear();
	for(int i=1;i<=m;i++){
		int from,to,dist;
		scanf("%d%d%d",&from,&to,&dist);
		int nfrom = min(from,to),	nto = max(from,to);
		pii p = make_pair(nfrom,nto);
		if(!st.count(p)){
			st.insert(p);
			mp[nfrom][nto] = dist;
		}else{
			if(mp[nfrom][nto] > dist)
				mp[nfrom][nto] = dist;
		}
	}
	for(set<pii>::iterator it=st.begin();it!=st.end();it++){
		int from = (*it).first,	to = (*it).second;
		mp[to][from] = mp[from][to];
	}
	Floyd();
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)
			printf("%d ",mp[i][j]);
		printf("\n");
	}
	return 0;
}