Bzoj 3445(最短路+枚举)

bzoj 3445
Luogu免权限地址
先对原图进行一次最短路，然后记录最短路上的边，然后枚举每一条边加倍，进行最短路，取每次最短路的答案减去原图最短路即可

#include<cstdio>  
#include<algorithm>  
#include<cstring>
#include<queue>  
#define ms(i,j) memset(i,j, sizeof i);
using namespace std;
const int MAXN = 100 + 10, INF = 100000000;
struct node
{
    int dis, no;
    bool operator < (const node &b) const {return dis>b.dis;}
};
struct edge
{
    int u,v;
}E[5000+100];
int cnt = 0;
int G[MAXN][MAXN];
int dis[MAXN];
int vi[MAXN];
int pre[MAXN];
int n,m;
int ans = 0, oa;
void fdij()
{
    priority_queue<node> pq;
    for (int i=1;i<=n;i++) dis[i] = INF, vi[i] = false, pre[i] = 0; dis[1] = 0;
    pq.push((node){0, 1});
    while (!pq.empty())
    {
        node p = pq.top(); pq.pop();
        if (vi[p.no]) continue;
        vi[p.no] = true;
        for (int i=1;i<=n;i++)
        if (G[i][p.no]!=INF)
        if (dis[p.no]+G[p.no][i]<dis[i])
        {
            dis[i] = dis[p.no]+G[p.no][i];
            pre[i] = p.no;
            pq.push((node){dis[i], i});
        }
    }
    oa = ans = dis[n];
}
void dij()
{
    priority_queue<node> pq;
    for (int i=1;i<=n;i++) dis[i] = INF, vi[i] = false; dis[1] = 0;
    pq.push((node){0, 1});
    while (!pq.empty())
    {
        node p = pq.top(); pq.pop();
        if (vi[p.no]) continue;
        vi[p.no] = true;
        for (int i=1;i<=n;i++)
        if (G[i][p.no]!=INF)
        if (dis[p.no]+G[p.no][i]<dis[i])
        {
            dis[i] = dis[p.no]+G[p.no][i];
            pq.push((node){dis[i], i});
        }
    }
    ans = max(ans, dis[n]);
}
int main()  
{  
    scanf("%d%d", &n, &m);
    for (int i=1;i<=n;i++)
    for (int j=1;j<=n;j++) G[i][j] = INF;
    for (int i=1;i<=m;i++)
    {
        int x,y,z;
        scanf("%d%d%d", &x, &y, &z);
        G[x][y] = G[y][x] = z;
    }
    fdij();
    int now = n;
    while (pre[now]!=0)
    {
        E[++cnt].u = now;
        E[  cnt].v = pre[now];
        now = pre[now];
    }
    for (int i=1;i<=cnt;i++)
    {
        int x = E[i].u;
        int y = E[i].v;
        G[x][y] *= 2;
        G[y][x] *= 2;
        dij();
        G[x][y] /= 2;
        G[y][x] /= 2;
    }
    printf("%d\n", ans-oa);
    return 0;  
}