Codeforces Round #287 (Div. 2) E. Breaking Good

题意 在一个稀疏图里找到从1号节点到n号节点的最短路，边权均为1，但路分好坏，在找到最短路的同时，还希望经过的好路最多。在最短路径上的坏路需要修好，不在最短路径上的好路要炸毁，问你最后在找到最短路的同时，需要修好和炸毁的路的总和是多少，并给出具体是哪些边。

思路 A*搜索，优先级1是经过的边数，优先级2是经过的坏路数，用最小值堆。就是要注意，需要保存所有搜过的状态，以便回溯找到那些最短路上的边。复杂度O(n log n)。

其实写完，才发现这不就是优先队列优化的Dijkstra的一点变形.....好吧，洞察力太次了...哎，比赛的时候邻接表没好好写，本来想节省时间的....结果最后发现不好好写，复杂度就退化了....然后又浪费时间改....最后没来及调完....泪目

#include <iostream>#include <cstdio>#include <cmath>#include <cstdio>#include <vector>#include <queue>#include <algorithm>#include <cstring>#include <string>#include <cstdlib>#include <map>using namespace std;#define I64_MAX 9223372036854775807 typedef long long ll;const double pi=acos (-1.0);const double eps=1e-8 ;//const ll INF=(I64_MAX)/2;//#pragma comment(linker, "/STACK:102400000,102400000")const int inf=0x3f3f3f3f ;#define maxx(a) memset(a, 0x3f, sizeof(a))#define zero(a) memset(a, 0, sizeof(a))#define FILL(a,b) memset(a, b, sizeof(a))#define REP(i,a,b) for(i=a;i<b;i++)#define rep(i,n) REP(i,0,n)#define srep(i,n) for(i = 1;i <= n;i ++)#define snuke(c,itr) for( __typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)#define MP make_pair#define fi first#define se secondtypedef pair <int, int> PII;typedef pair <ll, ll> PX;typedef pair<int,ll> PIL;#define MAX 100005 int n,m;int mark[MAX];int marke[MAX<<1];struct edge{int u,v,z;edge(int u = 0,int v= 0,int z =0 ):u(u),v(v),z(z){}}e[MAX*2];vector<int> g[MAX];struct node{int v;int p1,p2;int pre;node(int v=0,int p1=0,int p2=0,int pre = -1):v(v),p1(p1),p2(p2),pre(pre){}};node nodeSet[MAX];struct cmp1{bool operator() (node x,node y){return x.p1 > y.p1 || (x.p1 == y.p1 && x.p2 > y.p2);}};priority_queue<node,vector<node>,cmp1> q;int main (){    // freopen("E:\\input.txt" ,"r", stdin);     // freopen ("E:\\out.txt","w",stdout);int i,j;scanf("%d%d",&n,&m);for(i=0;i<m;i++){int t1,t2,t3;scanf("%d%d%d",&t1,&t2,&t3);e[i*2] = edge(t1,t2,t3);e[i*2+1] = edge(t2,t1,t3);g[t1].push_back(2*i);g[t2].push_back(2*i+1);}q.push(node(1,1,0,-1));node tmp;while(q.size() > 0){tmp = q.top();q.pop();if(mark[tmp.v] == 1)continue;mark[tmp.v] = 1;nodeSet[tmp.v] = tmp;if(tmp.v == n)break;for(i=0;i<g[tmp.v].size();i++){if(mark[e[g[tmp.v][i]].v] == 1)continue;int p,zz;p = tmp.p1 + 1;if(e[g[tmp.v][i]].z == 0)zz = tmp.p2+1;elsezz = tmp.p2;q.push(node(e[g[tmp.v][i]].v,p,zz,g[tmp.v][i]));//cout<<g[tmp.v][i]<<endl;}}//cout<<tmp.p1<<'\n';int num = 0;i = tmp.pre;while(i != -1){marke[i] = 1;i = nodeSet[e[i].u].pre;}for(i=0;i<2*m;i+=2){if(marke[i] == 1 || marke[i+1] == 1){if(e[i].z == 0)num++;}else{if(e[i].z == 1)num++;}}cout<<num<<'\n';for(i=0;i<2*m;i+=2){if(marke[i] == 1 || marke[i+1] == 1){if(e[i].z == 0)cout<<e[i].u<<' '<<e[i].v<<" 1\n";}else{if(e[i].z == 1)cout<<e[i].u<<' '<<e[i].v<<" 0\n";}}return 0;}