题目链接
题意
P A B X表示A在B点的北边的准确距离为X
V A B表示A在B的北边,但是具体的距离不确定,但是距离一定大于1.
询问是否存在一种情况使N个据点满足之前的条件。
思路
两个点的距离确定是x,相当于
a - b >= x
&&
a - b <= x
另外一个条件就是
a - b >= 1
通过这两个信息建边,可以求最短路也可以最长路,区别就是,最短路判断负环,最长路判断正环
PS:
坑死我了,边最多是20w + 1000个,我数组开小了,我开了11w,它提示我TLE,服了
CODE
#include <queue>
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#pragma warning (disable:4996)
#pragma warning (disable:6031)
#define mem(a, b) memset(a, b, sizeof a)
using namespace std;
const int N = 220000;
int head[N], nex[N], to[N], edge[N], cnt, _cnt[N], d[N];
bool v[N];
int n, m;
struct SPFA {
SPFA() {
mem(head, -1);
mem(nex, -1);
mem(_cnt, 0);
mem(d, 0x3f);
mem(v, false);
cnt = 0;
}
void init() {
SPFA();
}
void add(int a, int b, int c) {
++cnt;
to[cnt] = b;
edge[cnt] = c;
nex[cnt] = head[a];
head[a] = cnt;
}
bool spfa() {
queue<int> q;
q.push(0);
d[0] = 0;
v[0] = 1;
_cnt[0]++;
while (!q.empty()) {
int t = q.front();
q.pop();
v[t] = 0;
for (int i = head[t]; i != -1; i = nex[i]) {
int y = to[i];
int dist = edge[i];
if (d[y] > d[t] + dist) {
d[y] = d[t] + dist;
if (!v[y]) {
v[y] = 1;
q.push(y);
_cnt[y]++;
if (_cnt[y] > n) {
return false;
}
}
}
}
}
return true;
}
};
int main()
{
SPFA sp;
while (~scanf("%d %d", &n, &m)) {
sp.init();
int a, b, c;
char ch;
for (int i = 0; i < m; i++) {
getchar();
ch = getchar();
if (ch == 'V') {
scanf("%d %d", &a, &b);
sp.add(a, b, -1);
}
else {
scanf("%d %d %d", &a, &b, &c);
sp.add(b, a, c);
sp.add(a, b, -c);
}
}
for (int i = 1; i <= n; i++) {
sp.add(0, i, 0);
}
if (sp.spfa()) {
puts("Reliable");
}
else {
puts("Unreliable");
}
}
return 0;
}