HDU 4889 SPFA_killer 多校练习赛3

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题目意思很简单。。

cha 掉它程序，数据不能是负环。。

= = 那个程序用了SPFA的两个优化，竟然被负权边给干掉了。。下次我开优化的时候。。得考虑用优先队列处理负权边了。。不然就被cha了。。

SPFA_killer.

//tpl//ipqhjjybj_tpl.h//header.h#include <cstdio>#include <cstdlib>#include <map>#include <set>#include <algorithm>#include <cstring>#include <iostream>#include <vector>#include <string>#include <queue>#include <assert.h>#define mp(x,y) make_pair(x,y)#define pii pair<int,int>#define pLL pair<long long ,long long>#define rep(i,j,k) for(int i = j; i < k;i++)#define repn(i,j,k) for(int i = j; i <= k;i++)using namespace std;const int INF = 0x3f3f3f3f;int c;void print(){int n = 51;printf("%d %d\n",n,75);int ind = n;int w1 = 0 , w2 = -2;int cnt = 0;for(int i=n-2;i >= 1;i -=2){printf("%d %d %d\n",i,i+1,0);printf("%d %d %d\n",i+1,i+2,w2);printf("%d %d %d\n",i,i+2,w1);w1 = w2;w2 *= 2;cnt+=3;}//printf("cnt=%d n+2=%d\n",cnt,n+2);assert(cnt == 75);}int main(){while(scanf("%d",&c)!=EOF){print();}return 0;}

如果用 priority_queue。。真不知道怎样的数据能卡掉它。求高手卡掉下面的程序。。

//tpl//ipqhjjybj_tpl.h//header.h#include <cstdio>#include <cstdlib>#include <map>#include <set>#include <algorithm>#include <cstring>#include <iostream>#include <vector>#include <string>#include <queue>#include <assert.h>#define mp(x,y) make_pair(x,y)#define pii pair<int,int>#define pLL pair<long long ,long long>#define rep(i,j,k) for(int i = j; i < k;i++)#define repn(i,j,k) for(int i = j; i <= k;i++)using namespace std;#define C 23333333int main() {int n,m;cin >> n >> m;vector<pair<int,int> > edges[111];for(int i = 0;i < m;i++) {int x,y,w;cin >> x >> y >> w;edges[x].push_back(make_pair(y,w));}//deque<int> q;priority_queue<pii > q;vector<long long> dist(n+1, ~0ULL>>1);vector<bool> inQueue(n+1, false);dist[1] = 0; q.push(mp(0,1)); inQueue[1] = true;int doge = 0;while(!q.empty()) {int x = q.top().second; q.pop();if(doge++ > C) {puts("doge");return 233;}for(vector<pair<int,int> >::iterator it = edges[x].begin();it != edges[x].end();++it) {int y = it->first;int w = it->second;if(dist[y] > dist[x] + w) {dist[y] = dist[x] + w;if(!inQueue[y]) {inQueue[y] = true;q.push(mp(dist[y],y));}}inQueue[x] = false;}}printf("%d\n",dist[n]);}

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