hdu-6181 Two Paths次短路

题意：求次短路。

思路：求次短路

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#define siz 200005#define LL long longnamespace fastIO {    #define BUF_SIZE 100000    //fread -> read    bool IOerror = 0;    inline char nc() {        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;        if(p1 == pend) {            p1 = buf;            pend = buf + fread(buf, 1, BUF_SIZE, stdin);            if(pend == p1) {                IOerror = 1;                return -1;            }        }        return *p1++;    }    inline bool blank(char ch) {        return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';    }    inline void read(int &x) {        char ch;        while(blank(ch = nc()));        if(IOerror)            return;        for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');    }    #undef BUF_SIZE};using namespace fastIO;using namespace std;const LL INF = 1e18;struct qnode{    int v;    LL c;    qnode(int _v = 0,LL _c = 0) : v(_v),c(_c){}    bool operator < (const qnode &r) const{        return c>r.c;    }};struct Edge{    int v;    LL cost;    Edge(int _v = 0,LL _cost = 0) : v(_v),cost(_cost){}};int n,m;vector<Edge>E[siz];bool vis[siz];LL dist1[siz];LL dist2[siz];void addedge(int u,int v,LL w){    E[u].push_back(Edge(v,w));}void dijstra(int n,int start){    //memset(vis,0,sizeof(vis));    for(int i=1;i<=n;i++){        dist1[i] = INF;        dist2[i] = INF;    }    priority_queue<qnode> que;    while(!que.empty()) que.pop();    dist1[start] = 0;    que.push(qnode(start,0));    qnode tmp;    while(!que.empty()){        tmp = que.top();        que.pop();        int u = tmp.v;        LL d = tmp.c;        if(dist2[u] < d) continue;        for(int i = 0;i<E[u].size();i++){            int v = E[u][i].v;            LL cost = E[u][i].cost;            LL d2 = d + cost;            if(dist1[v] > d2){                swap(d2,dist1[v]);                que.push(qnode(v,dist1[v]));            }            if(dist2[v] > d2&&dist1[v] <d2){                dist2[v] = d2;                que.push(qnode(v,dist2[v]));            }        }    }    printf("%lld\n",dist2[n]);}void solve(){    dijstra(n,1);}int main(){    int T;    //scanf("%d",&T);    read(T);    while(T--){        //scanf("%d%d",&n,&m);        read(n);        read(m);        for(int i=0;i<=n;i++){            E[i].clear();        }        for(int i = 1;i<=m;i++){            int u,v;            int w;            //scanf("%d%d%lld",&u,&v,&w);            read(u);            read(v);            read(w);            addedge(u,v,w);            addedge(v,u,w);        }        solve();    }    return 0;}

保存一波大佬的A*求k短路模板。

A*算法模板在次，点击传送