UESTCoj 265 War（Floyd最短路）

题目描述：Your country is now involved in a war! In the front, there are N positions
between which you can transfer goods. Unfortunately the enemies will attack you and destroy some positions. When transferring, you cannot pass through a position which has been destroyed. As the commander of the army, you want to know the minimum distance
between some position u and v,
after some positions have been destroyed.

Floyd算法求最短路的原理是枚举每个点做为中点，更新起点和终点的最短距离。

按照题意中的意思是要拆除一些点后再求最短路，我们可以倒着去做认为是新建一些点后再去求最短路，对于每个新建点k，我们只要以k为中点更新一下最短路就行了，更新的过程是O(n^2)的。然后最多新建n个点，所以整个算法复杂度为O(n^3)。

PS：我觉得暴力SPFA应该可以过的，然后就T了。

#pragma warning(disable:4996)
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

int dis[205][205];
bool exist[205];
vector<int>ans;
int query[10005], x[10005], y[10005];

int n, m;

int main(){
int t; scanf("%d", &t);
while (t--){
scanf("%d %d", &n, &m);
memset(dis, 0x3f, sizeof dis);
memset(exist, true, sizeof exist);
for (int i = 1; i <= n; i++)dis[i][i] = 0;
for (int i = 0; i < m; i++){
int u, v, c; scanf("%d %d %d", &u, &v, &c);
dis[u][v] = dis[v][u] = min(dis[u][v], c);
}

int q; scanf("%d", &q);
for (int i = 1; i <= q; i++){
int op; scanf("%d", &op);
if (op == 0){
int u, v; scanf("%d %d", &u, &v);
query[i] = 0;
x[i] = u;
y[i] = v;
}
else if (op == 1){
int u; scanf("%d", &u);
exist[u] = false;
query[i] = 1;
x[i] = u;
}
}
//Floyd求出最短路
for (int k = 1; k <= n; k++){
if (!exist[k])continue;
for (int i = 1; i <= n; i++){
for (int j = 1; j <= n; j++){
dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
}
}
}

ans.clear();
for (int i = q; i >= 1; i--){
if (query[i] == 0){
if (exist[x[i]] && exist[y[i]]){
int tmp = dis[x[i]][y[i]];
tmp = tmp == dis[0][0] ? -1 : tmp;
ans.push_back(tmp);
}
else ans.push_back(-1);
}
//加点
else if (query[i] == 1){
int k = x[i];
exist[k] = true;
for (int i = 1; i <= n; i++){
for (int j = 1; j <= n; j++){
dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
}
}
}
}
for (int i = (int)ans.size() - 1; i >= 0; i--){
printf("%d\n", ans[i]);
}
puts("");
}
return 0;
}