题目链接:点我啊╭(╯^╰)╮
题目大意:
有多种汇币,汇币之间可以交换,这需要手续费,当你用100A币交换B币时,A到B的汇率是29.75,手续费是0.39,那么你可以得到(100 - 0.39) * 29.75 = 2963.3975 B币。问s币的金额经过交换最终得到的s币金额数能否增加
解题思路:
判断正环,这里可以用Bellman-Ford,也能用spfa,这里给出一种类似Bellman-Ford算法的算法,和正确的spfa算法
代码思路:
感觉像是floyd,但感觉又不是,看出来的人麻烦和我说下。。注意用spfa的时候不需要判断入队次数,只需要判断起点的更新状态能否大于初始状态即可
核心:正负权回路的判断
Bellman-Ford:
include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n, m, st;
double v, ori[105], re[105], k1[105][105], k2[105][105];
bool floyed() {
for(int k=0; k<n; k++) //插入k点
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
re[j]= max(re[j], (re[i] - k2[i][j]) * k1[i][j]);
for(int i=0; i<n; i++)
ori[i] = re[i];
for(int k=0; k<n; k++) //插入k点
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
re[j]= max(re[j], (re[i] - k2[i][j]) * k1[i][j]);
for(int i=0; i<n; i++)
if(re[i]>ori[i])
return true;
return false;
}
void init() {
scanf("%d%d%d%lf", &n, &m, &st, &v);
re[st-1] = v;
int x, y;
double a, b, c, d;
for(int i=0; i<m; i++) {
scanf("%d %d %lf %lf %lf %lf", &x, &y, &a, &b, &c, &d);
k1[x-1][y-1] = a; k2[x-1][y-1] = b;
k1[y-1][x-1] = c; k2[y-1][x-1] = d;
}
}
int main() {
init();
if(floyed()) puts("YES");
else puts("NO");
}
SPFA:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 200010;
const int MAXM = 500010;
struct EDGE {
int next;
int to;
double k1, k2;
};
EDGE edge[MAXM];
int n, m, st, cnt;
double v, dis[MAXN];
int head[MAXN], num[MAXN];
bool vis[MAXN];
queue<int> Q;
void Add(int u, int v, double k1, double k2) {
edge[++cnt].next = head[u];
edge[cnt].to = v;
edge[cnt].k1 = k1;
edge[cnt].k2 = k2;
head[u] = cnt;
}
void read() {
int x, y;
double a, b, c, d;
scanf("%d%d%d%lf", &n, &m, &st, &v);
for(int i=1; i<=m; i++) {
scanf("%d%d%lf%lf%lf%lf", &x, &y, &a, &b, &c, &d);
Add(x, y, a, b);Add(y, x, c, d);
}
}
bool SPFA(int x) {
while(!Q.empty()) Q.pop();
cnt = 0;
dis[x] = v;
num[x] = 1;
Q.push(x);
vis[x] = true;
while(!Q.empty()) {
int k = Q.front();
Q.pop();
vis[k] = false;
for(int i=head[k]; i!=0; i=edge[i].next) {
int j = edge[i].to;
double k1 = edge[i].k1;
double k2 = edge[i].k2;
if(dis[j] < (dis[k]-k2)*k1) {
dis[j] = (dis[k]-k2)*k1;
num[j] = num[k]+1;
//if(num[j]>n) return 1; //判断负环
if(!vis[j]) {
Q.push(j);
vis[j] = true;
}
}
if(dis[x]>v) return 1;
}
}
return 0;
}
int main() {
memset(vis, 0, sizeof(vis));
memset(num, 0, sizeof(num));
memset(head, 0, sizeof(head));
read();
SPFA(st) ? printf("YES\n") : printf("NO\n");
}