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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;
}
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原文地址:http://www.cnblogs.com/NanoApe/p/4330725.html