题目链接：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=4162&mosmsg=Submission+received+with+ID+26583689

题目要求所有点对间的最短路之和，因为是稀疏图，floyd 和 dijstra 差不多，如果枚举 m 条边每次求一遍答案，时间无法承受

考虑最短路树，如果源点确定，那么只有修改在最短路树上的边，才会改变源点到其他点的最短距离，而这样的边只有 \(n-1\) 条

所以枚举所有在最短路树上的边即可，要注意的细节就是因为是无向图，删边时要将反向边也删掉

时间复杂度 \(O(n^2mlogn)\)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 110;
const int INF = 0x3f3f3f3f;

int n, m, L;
int u[1010], v[1010], w[1010];

int h[maxn], cnt = 1;
struct E{
	int from, to, cost, next;
}e[maxn * 10 * 2];
void add(int u, int v, int w){
	e[++cnt].to = v;
	e[cnt].from = u;
	e[cnt].cost = w;
	e[cnt].next = h[u];
	h[u] = cnt;
}

ll ans1, ans2[3010];
int d[maxn], pre[maxn], vis[3010];
void dij(int S, int ban){
	memset(pre, 0, sizeof(pre));
	memset(d, 0x3f, sizeof(d));
	d[S] = 0;
	priority_queue<pii, vector<pii>, greater<pii> > q;
	q.push(pii(d[S], S));
	
	while(!q.empty()){
		pii p = q.top(); q.pop();
		int u = p.second;
		if(p.first != d[u]) continue;
		
		for(int i = h[u] ; i != -1 ; i = e[i].next){
			int v = e[i].to;
			if(i == ban || i == (ban ^ 1)) continue;
			if(d[v] > d[u] + e[i].cost){
				pre[v] = i; 
				d[v] = d[u] + e[i].cost;
				q.push(pii(d[v], v));
			}
		}
	}
}

void solve(int u){
	int res1 = 0;
	dij(u, -1); // 求出最短路树 
	for(int i = 1 ; i <= n ; ++i){
		ans1 += d[i] == INF ? L : d[i];	
	}

	for(int i = 1 ; i <= n ; ++i){ // 枚举删除哪条边 
		if(!pre[i]) continue; 
		vis[pre[i]] = 1; 
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < ‘0‘ || ch > ‘9‘){ if(ch == ‘-‘) f = -1; ch = getchar(); } while(ch >= ‘0‘ && ch <= ‘9‘){ s = s * 10 + ch - ‘0‘; ch = getchar(); } return s * f; }

int main(){
	while(scanf("%d%d%d", &n, &m, &L) == 3){
		ans1 = 0;
		memset(ans2, 0, sizeof(ans2));
		memset(vis, 0, sizeof(vis));
		memset(h, -1, sizeof(h)); cnt = 1;
		for(int i = 1 ; i <= m ; ++i){
			scanf("%d%d%d", &u[i], &v[i], &w[i]);
			add(u[i], v[i], w[i]); add(v[i], u[i], w[i]);
		}
		for(int i = 1 ; i <= n ; ++i){
			solve(i);
		}
		
		for(int i = 2 ; i <= cnt ; i += 2){
			if(vis[i] || vis[i^1]){
				for(int j = 1 ; j <= n ; ++j){
					dij(j, i);
					for(int k = 1 ; k <= n ; ++k){
						ans2[i] += d[k] == INF ? L : d[k];
					}
				}
			}
		}
		
		ll res2 = 0;
		for(int i = 1 ; i <= cnt ; ++i) res2 = max(res2, ans2[i]);
		printf("%lld %lld\n", ans1, res2);
	}
	return 0;
}

UVa 1416 - Warfare And Logistics (最短路树)