题目连接
- 题意：
给定一个无向图，每个边有两个属性，长度和一个字母‘L‘，’O‘，‘V’，‘E‘中的一个。从1点开始到达n点，每次必须按照L -> O -> V -> E -> ... -> E的顺序，到达终点时候必须经过E边
- 分析：
对于这种对边的限制，比较简单的方法是将一个点拆成若干个点。因为经过’L‘到达点p的状态和经过‘O‘到达点p的状态时不一样的，第一个之后只能经过’O‘边，而第二个只能经过‘V‘边，所以经过不同的边到达同一个点的时候对应的状态应该分开，也就是将点拆分成四个点，分别表示经过四种边到达p点。
- 注意：
图可以有自环，也可以只有一个点
路径必须至少有一个LOVE
路径长度尽可能小，长度若相等，那么LOVE的数量尽可能多
Dijkstra方法：（一个点的时候直接特判，避免不必要的麻烦）
const LL INF = 1e18;
const int MAXV = 10000;
struct Edge
{
LL from, to, dist;
};
struct HeapNode
{
LL d, u, num;
bool operator < (const HeapNode& rhs) const
{
return d > rhs.d;
}
};
struct Dijkstra
{
int n; //n:点数 m:临时变量
vector<Edge> edges; //存储所有的边
vector<int> G[MAXV];//每个点的所有相邻边序号
bool done[MAXV]; // 是否已永久标号
LL d[MAXV]; // s起点到各个点的距离
LL num[MAXV];
void init(int n)
{
this->n = n;
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int dist)
{
G[from].push_back(edges.size());
edges.push_back((Edge) { from, to, dist });
}
void dijkstra(int s)
{
priority_queue<HeapNode> Q;
for(int i = 0; i < n; i++) d[i] = INF;
CLR(num, 0);
d[s] = 0;
memset(done, 0, sizeof(done));
Q.push((HeapNode) { 0, s, 0 });
while(!Q.empty())
{
HeapNode x = Q.top();
Q.pop();
int u = x.u;
if(done[u]) continue;
done[u] = true;
for(int i = 0; i < G[u].size(); i++)
{
Edge& e = edges[G[u][i]];
if (d[e.to] == d[u] + e.dist)
num[e.to] = max(num[e.to], num[x.u] + 1);
if(d[e.to] > d[u] + e.dist)
{
d[e.to] = d[u] + e.dist;
num[e.to] = num[x.u] + 1;
Q.push((HeapNode) { d[e.to], e.to, num[e.to] });
}
}
}
}
} dij;
LL chk[4];
int main()
{
int T;
RI(T);
FE(kase, 1, T)
{
REP(i, 4) chk[i] = INF;
int n, m, u, v, d, op;
char type;
RII(n, m);
dij.init(n << 2);
REP(i, m)
{
scanf("%d%d%d %c", &u, &v, &d, &type);
u--; v--;
if (type == 'L') op = 0;
else if (type == 'O') op = 1;
else if (type == 'V') op = 2;
else op = 3;
chk[op] = min(chk[op], (LL)d);
dij.AddEdge(u + (op + 3) % 4 * n, v + op * n, d);
dij.AddEdge(v + (op + 3) % 4 * n, u + op * n, d);
}
printf("Case %d: ", kase);
if (n == 1)
{
REP(i, 4)
if (chk[i] == INF)
{
puts("Binbin you disappoint Sangsang again, damn it!");
goto end;
}
printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %d LOVE strings at last.\n"
, chk[0] + chk[1] + chk[2] + chk[3], 1);
end:;
}
else
{
dij.dijkstra(3 * n);
if (dij.d[4 * n - 1] == INF)
puts("Binbin you disappoint Sangsang again, damn it!");
else
{
printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %I64d LOVE strings at last.\n"
, dij.d[4 * n - 1], dij.num[4 * n - 1] / 4);
}
}
}
return 0;
}
spfa方法：（一个点也是特判，加点方式与Dijkstra相同）
const LL INF = 1e18;
const int MAXV = 10000;
struct Edge
{
int from, to, dist;
};
struct SPFA
{
int n;
LL d[MAXV];
int num[MAXV];
vector<Edge> edges;
vector<int> G[MAXV];
bool inq[MAXV];
void init(int n)
{
this->n = n;
edges.clear();
REP(i, n)
G[i].clear();
}
void AddEdge(int from, int to, int dist)
{
G[from].push_back(edges.size());
edges.push_back((Edge) {from, to, dist});
}
void spfa(int s)
{
queue<int> q;
CLR(inq, false);
CLR(num, 0);
REP(i, n) d[i] = INF;
d[s] = 0;
q.push(s); inq[s] = true;
while (!q.empty())
{
int u = q.front();
q.pop(); inq[u] = false;
REP(i, G[u].size())
{
Edge& e = edges[G[u][i]];
if (d[e.to] == d[u] + e.dist && num[u] + 1 > num[e.to])
{
num[e.to] = num[u] + 1;
if (!inq[e.to])
{
q.push(e.to);
inq[e.to] = true;
}
}
if(d[e.to] > d[u] + e.dist)
{
d[e.to] = d[u] + e.dist;
num[e.to] = num[u] + 1;
if (!inq[e.to])
{
q.push(e.to);
inq[e.to] = true;
}
}
}
}
}
} spfa;
LL chk[4];
int main()
{
int T;
RI(T);
FE(kase, 1, T)
{
REP(i, 4) chk[i] = INF;
int n, m, u, v, d, op;
char type;
RII(n, m);
spfa.init(n << 2);
REP(i, m)
{
scanf("%d%d%d %c", &u, &v, &d, &type);
u--; v--;
if (type == 'L') op = 0;
else if (type == 'O') op = 1;
else if (type == 'V') op = 2;
else op = 3;
chk[op] = min(chk[op], (LL)d);
spfa.AddEdge(u + (op + 3) % 4 * n, v + op * n, d);
spfa.AddEdge(v + (op + 3) % 4 * n, u + op * n, d);
}
printf("Case %d: ", kase);
if (n == 1)
{
REP(i, 4)
if (chk[i] == INF)
{
puts("Binbin you disappoint Sangsang again, damn it!");
goto end;
}
printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %d LOVE strings at last.\n"
, chk[0] + chk[1] + chk[2] + chk[3], 1);
end:;
}
else
{
spfa.spfa(3 * n);
if (spfa.d[4 * n - 1] == INF)
puts("Binbin you disappoint Sangsang again, damn it!");
else
{
printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %d LOVE strings at last.\n"
, spfa.d[4 * n - 1], spfa.num[4 * n - 1] / 4);
}
}
}
return 0;
}
顺便弄一些自己的测试数据，方便查错。
4 4
1 2 1 L
2 1 1 O
1 3 1 V
3 4 1 E
ans:4, 1
4 4
1 2 1 L
2 3 1 O
3 4 1 V
4 1 1 E
ans:no
12 12
1 5 10 L
5 6 10 O
6 7 10 V
7 12 10 E
1 2 1 L
2 3 1 O
3 4 1 V
4 8 1 E
8 9 1 L
9 10 1 O
10 11 1 V
11 12 33 E
ans:40, 2
12 12
1 5 10 L
5 6 10 O
6 7 10 V
7 12 10 E
1 2 1 L
2 3 1 O
3 4 1 V
4 8 1 E
8 9 1 L
9 10 1 O
10 11 1 V
11 12 34 E
ans:40, 1
1 4
1 1 1 L
1 1 1 O
1 1 1 V
1 1 1 E
ans:4, 1
2 8
1 1 2 L
1 1 1 O
1 1 1 V
1 1 1 E
1 2 3 L
2 1 1 O
1 2 1 V
2 1 1 E
ans:5, 1
1 3
1 1 1 L
1 1 1 O
1 1 1 E
ans:no
11 11
1 2 1 L
2 3 1 O
3 4 348 V
4 11 1000 E
1 5 50 L
5 6 50 O
6 7 50 V
7 8 50 E
8 9 50 L
9 10 50 O
10 4 50 V
ans:1350 2
As long as Binbin loves Sangsang
时间： 2024-10-15 18:16:36