UVA 1416 Warfare And Logistics 最短路树

对每个点求最短路，同时求出最短路树。

枚举每条边，如果这条边在最短路树上，那么删掉这条边就需要重新计算最短路，否则不需要。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\in1.txt","r",stdin);
    freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}

struct Edge
{
    int from,to,dist;
};
struct HeapNode
{
    int d,u;
    bool operator < (const HeapNode &ant ) const
    {
        return d>ant.d;
    }
};

const int maxn = 105;
int n,m;
vector<int> g[maxn];
vector<Edge> edge;
int done[maxn];
void init()
{
    for(int i=1;i<=n;i++)
        g[i].clear();
    edge.clear();
}
void add(int u,int v,int w)
{
    Edge e=(Edge){u,v,w};
    edge.push_back(e);
    g[u].push_back(edge.size()-1);
}
void dijksta(int s,int *d,int *p,int eid)
{
    for(int i=1;i<=n;i++)
        d[i]=INF;
    memset(done,0,sizeof(done));
    priority_queue<HeapNode> q;

    p[s]=-1;
    d[s]=0;
    q.push((HeapNode){0,s});

    while(!q.empty())
    {
        HeapNode x=q.top();q.pop();
        int u=x.u;
        if(done[u])continue;
        done[u]=1;
        for(int i=0;i<g[u].size();i++)
        {
            if(g[u][i]==eid||g[u][i]==(eid^1))continue;
            Edge &e=edge[g[u][i]];
            if(d[e.to]>d[u]+e.dist)
            {
                d[e.to]=d[u]+e.dist;
                p[e.to]=g[u][i];
                q.push((HeapNode){d[e.to],e.to});
            }
        }
    }
}

int d[maxn][maxn],p[maxn][maxn];
int L;
long long sigma(int u)
{
    long long c=0;
    for(int i=1;i<=n;i++)
        c=c+(d[u][i]==INF?L:d[u][i]);
    return c;
}
long long solve()
{
    long long maxc=-1;
    for(int i=0;i<2*m;i+=2)
    {
        long long c=0;
        for(int j=1;j<=n;j++)
        {
            if(p[j][edge[i].to]==i||p[j][edge[i+1].to]==i+1)
            {
                dijksta(j,d[j],p[j],i);
                c+=sigma(j);
                dijksta(j,d[j],p[j],m*2);
            }else
                c+=sigma(j);
        }
        maxc=max(maxc,c);
    }
    return maxc;
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&L)!=EOF)
    {
        init();
        for(int i=0;i<m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        for(int i=1;i<=n;i++)
            dijksta(i,d[i],p[i],2*m);
        long long c=0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                c=c+(d[i][j]==INF?L:d[i][j]);
        long long c1=solve();
        printf("%lld %lld
",c,c1);
    }
    return 0;
}