BZOJ 1624: [Usaco2008 Open] Clear And Present Danger 寻宝之路

最短路问题
只要把所有点之间的最短路跑出来就好了
Floyd O(n3)
然而我用了斐(zi)波(dai)那(S)契(T)堆(L)优(lie)化的dijkstra
明明理论复杂度是 O(n2log2n) 的说
哭晕在厕所

---

#include
#include
#include
#include
#define g getchar()
#define p pair
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
inline ll read(){
    ll x=0,f=1;char ch=g;
    for(;ch<'0'||ch>'9';ch=g)if(ch=='-')f=-1;
    for(;ch>='0'&&ch<='9';ch=g)x=x*10+ch-'0';
    return x*f;
}
inline void out(ll x){
    int a[25],wei=0;
    if(x<0)putchar('-'),x=-x;
    for(;x;x/=10)a[++wei]=x%10;
    if(wei==0){puts("0");return;}
    for(int j=wei;j>=1;--j)putchar('0'+a[j]);
    putchar('\n');
}
int n,m,a[10005],dis[105][105],d[105][105];
bool pd[105];
priority_queuevector
,greater
 >Q;
int clear(){
    for(;!Q.empty();Q.pop());
    memset(pd,0,sizeof(pd));
}
int main(){
//  freopen("","r",stdin);
//  freopen("","w",stdout);
    n=read();m=read();
    for(int i=1;i<=m;++i)a[i]=read();
    for(int i=1;i<=n;++i)
    for(int j=1;j<=n;++j)dis[i][j]=read();
    memset(d,inf,sizeof(d));
    for(int i=1;i<=n;++i){
        clear();
        d[i][i]=0;
        Q.push(make_pair(d[i][i],i));
        for(int k=1;kfor(;(!Q.empty())&&pd[Q.top().second];Q.pop());
            int u=Q.top().second;pd[u]=1;
            for(int j=1;j<=n;++j)if((!pd[j])&&d[i][u]+dis[u][j]int ans=0;
    for(int i=1;i1]];
    out(ans);
    return 0;
}