标签:
先处理成前缀和关系,然后可以很明显得看得出这是一个差分约束。那么就是最短路问题了。
顺便复习了一下SPFA加SLF优化是怎么写的,也学习到了另一个STL——Deque双向队列。
#include <cstdlib> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <fstream> #include <iostream> #include <deque> #define rep(i, l, r) for(int i = l; i <= r; i++) #define down(i, l, r) for(int i = l; i >= r; i--) #define N 123 #define M 2345 #define ll long long using namespace std; inline int read() { int x=0, f=1; char ch=getchar(); while (ch<‘0‘ || ch>‘9‘) { if (ch==‘-‘) f=-1; ch=getchar(); } while (ch>=‘0‘ && ch<=‘9‘) { x=x*10+ch-‘0‘; ch=getchar(); } return x*f; } struct edge{int y, z, n;} e[M]; int fir[N], en; int t, n, m, d[N], c[N], x, y, z; bool b[N], ans; deque <int> q; void AddE(int x, int y, int z) { en++, e[en].y = y, e[en].z = z, e[en].n = fir[x], fir[x] = en; en++, e[en].y = x, e[en].z = -z, e[en].n = fir[y], fir[y] = en; } void Init() { rep(i, 1, en) e[i] = e[0]; rep(i, 0, n) fir[i] = c[i] = 0; en = 0; ans = true; } int main() { t=read(); while (t--) { Init(); n=read(); m=read(); rep(i, 1, m) x=read(), y=read(), z=read(), AddE(x-1, y, z); rep(i, 0, n) d[i]=0, b[i]=1, c[i]=1, q.push_back(i); while (!q.empty()) { int x = q.front(); q.pop_front(), b[x]=0; if (c[x]>n+1) { ans=false; break; } int o = fir[x], y = e[o].y; while (o) { if (d[y] > d[x]+e[o].z) { d[y] = d[x]+e[o].z; if (!b[y]) b[y]=1, c[y]++, !q.empty()&&d[q.front()]>d[y] ? q.push_front(y) : q.push_back(y); } o=e[o].n, y=e[o].y; } } if (ans) printf("true\n"); else printf("false\n"); } return 0; }
标签:
原文地址:http://www.cnblogs.com/NanoApe/p/4330725.html