洛谷P1119 灾后重建 Floyd + 离线

https://www.luogu.org/problemnew/show/P1119

真是有故事的一题呢

半年前在宁夏做过一道类似的题，当时因为我的愚昧痛失了金牌。

要是现在去肯定稳稳的过，真是生不逢时。

简单的来说就是按照时间顺序一边建边一边查询，对于每次建边只要用floyd加中转点即可。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
inline int read(){int now=0;register char c=getchar();for(;!isdigit(c);c=getchar());
for(;isdigit(c);now=now*10+c-'0',c=getchar());return now;}
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
const double eps = 1e-9;
const int maxn = 210;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
const int maxm = 5e4 + 10;
int N,M,Q;
pair<int,int>P[maxn];
int dp[maxn][maxn];
struct Query{
    int x,y,t;
}query[maxm];
bool vis[maxn];
bool cmp(Query a,Query b){
    return a.t < b.t;
}
void add(int x){
    for(int i = 0; i < N ; i ++){
        if(i == x) continue;
        for(int j = 0; j < N ; j ++){
            if(j == x) continue;
            dp[i][j] = min(dp[i][j],dp[i][x] + dp[x][j]);
        }
    }
}
int main(){
    //freopen("C.in", "r", stdin);
    Sca2(N,M);
    for(int i = 0; i <= N - 1 ; i ++){
        Sca(P[i].first);
        P[i].second = i;
    }
    Mem(dp,0x3f);
    for(int i = 0; i <= N - 1; i ++) dp[i][i] = 0;
    for(int i = 1; i <= M ; i ++){
        int u,v,w; Sca3(u,v,w);
        dp[u][v] = dp[v][u] = w;
    }
    sort(P ,P + N);
    Sca(Q);
    int now = 0,cnt = 0;
    while(cnt < N && P[cnt].first == 0){
        vis[cnt] = 1;
        add(P[cnt++].second);
    } 
    for(int i = 1; i <= Q; i ++){
        Sca3(query[i].x,query[i].y,query[i].t);
        while(cnt < N && P[cnt].first <= query[i].t){
            vis[cnt] = 1;
            add(P[cnt++].second);
        }
        if(!vis[query[i].x] || !vis[query[i].y] || dp[query[i].x][query[i].y] == INF) Pri(-1);
        else Pri(dp[query[i].x][query[i].y]);
    }
    return 0;
}