UVALive 4080 Warfare And Logistics （最短路树）

很多的边会被删掉，需要排除一些干扰进行优化。

和UVA - 1279 Asteroid Rangers类似，本题最关键的地方在于，对于一个单源的最短路径来说，如果最短路树上的边没有改变的话，那么最短路肯定是不会变的，

所以只要枚举删掉最短路树上的边。这样的时间复杂度就能过了。

 

#include<bits/stdc++.h>
using namespace std;

const int maxn = 102, maxm = 2002;
int head[maxn], to[maxm], nxt[maxm],wei[maxm],ecnt;
int delta[maxm];

void addEdge(int u,int v,int w)
{
    to[ecnt] = v;
    nxt[ecnt] = head[u];
    wei[ecnt] = w;
    head[u] = ecnt++;
}

typedef pair<int,int> Node;
#define fi first
#define se second

int d[maxn],pa[maxn];
typedef long long ll;

void dijkstra(int n,int s,int L,bool flag,int e1,int e2)// e1 e2 边应该忽略
{
    priority_queue<Node,vector<Node>,greater<Node> > q;
    fill(d,d+n,L);
    d[s] = 0; q.push(Node(0,s)); pa[s] = -1;
    while(q.size()){
        Node x = q.top(); q.pop();
        int u = x.se;
        if(x.fi != d[u]) continue;
        for(int i = head[u]; ~i; i = nxt[i])if(i!=e1 && i!=e2) {
            int v = to[i];
            if(d[v] > d[u]+wei[i]){
                d[v] = d[u]+wei[i];
                if(flag) pa[v] = i; // 入边编号
                q.push(Node(d[v],v));
            }
        }
    }
}

ll cal(int n)
{
    ll ret = 0;
    for(int i = 0; i < n; i++){
        ret += d[i];
    }
    return ret;
}


void init()
{
    memset(head,-1,sizeof(head));
    ecnt = 0;
    memset(delta,0,sizeof(delta));
}

int main()
{
    //freopen("in.txt","r",stdin);
    int n,m,L;
    while(~scanf("%d%d%d",&n,&m,&L)){
        init();
         while(m--){
            int u,v,w; scanf("%d%d%d",&u,&v,&w);
            addEdge(--u,--v,w); addEdge(v,u,w);
        }
        ll ans1 = 0;
        vector<int> edges;
        for(int u = 0; u < n; u++){
            dijkstra(n,u,L,true,-1,-1);
            ll t  = cal(n);
            ans1 += t;
            for(int i = 0; i < n; i++){
                if(~pa[i]){
                    dijkstra(n,u,L,false,pa[i],pa[i]^1);
                    int eid = pa[i]&(~1);
                    if(!delta[eid]) edges.push_back(eid);
                    delta[eid] += cal(n)-t;
                }
            }
        }
        int MaxDelta = 0;
        for(int i = 0; i < (int)edges.size(); i++){
            MaxDelta = max(MaxDelta,delta[edges[i]]);
        }
        printf("%lld %lld
",ans1,ans1+MaxDelta);
    }
    return 0;
}