题目大意:幼儿园老师给小盆友们发糖果。有5种要求,问老师最少需要准备多少糖果。如不能满足,输出-1。
思路:裸地差分约束系统,但是正向加边会T,需要反向加边。
CODE:
#include <queue> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 400010 using namespace std; int points,asks; int head[MAX],total; int next[MAX],aim[MAX],length[MAX]; int f[MAX],showed_time[MAX]; bool v[MAX]; inline void Add(int x,int y,int len) { next[++total] = head[x]; aim[total] = y; length[total] = len; head[x] = total; } bool SPFA() { static queue<int> q; while(!q.empty()) q.pop(); q.push(0); memset(f,0xef,sizeof(f)); f[0] = 0; while(!q.empty()) { int x = q.front(); q.pop(); v[x] = false; for(int i = head[x]; i; i = next[i]) if(f[aim[i]] < f[x] + length[i]) { f[aim[i]] = f[x] + length[i]; if(!v[aim[i]]) { v[aim[i]] = true; q.push(aim[i]); showed_time[aim[i]] = showed_time[x] + 1; if(showed_time[aim[i]] > points + 1) return false; } } } return true; } int main() { cin >> points >> asks; for(int i = points; i; --i) Add(0,i,1); for(int flag,x,y,i = 1; i <= asks; ++i) { scanf("%d%d%d",&flag,&x,&y); if(flag == 1) Add(x,y,0),Add(y,x,0); if(flag == 2) Add(x,y,1); if(flag == 3) Add(y,x,0); if(flag == 4) Add(y,x,1); if(flag == 5) Add(x,y,0); } if(!SPFA()) { puts("-1"); return 0; } long long ans = 0; for(int i = 1; i <= points; ++i) ans += f[i]; cout << ans << endl; return 0; }
时间: 2024-10-06 00:31:01