无法找到 spoj MMINPAID 的错误答案测试用例
Unable to find a wrong answer testcase for spoj MMINPAID
问题link:http://www.spoj.com/problems/MMINPAID/
提交时获得 WA。
我使用 bfs 到达第 N 个节点并使用位掩码计算路径中节点的最小成本。然而,在 运行 大量测试用例并与公认的解决方案进行比较之后,无法找到失败的测试用例。
代码:
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int MAXN = 15, INF = 1 << 29;
struct node {
int c, p, r;
};
struct node data[MAXN][MAXN];
vector<int> g[MAXN];
int N, M, dist[MAXN][1 << 11];
int bfs() {
for (int i = 0; i < MAXN; i++)
for (int k = 0; k < (1 << 11); k++)
dist[i][k] = INF;
queue< pair< pair <int, int> , int > > q;
int v = 1, path = (1 << 1), cost = 0;
dist[v][path] = 0;
q.push(make_pair(make_pair(v, path), cost));
while (!q.empty()) {
int curv = q.front().first.first;
int curpath = q.front().first.second;
int curcost = q.front().second;
q.pop();
for (int i = 0; i < g[curv].size(); i++) {
int nv = g[curv][i];
int d1 = curcost + data[curv][nv].r;
int d2 = INF;
if (curpath & (1 << data[curv][nv].c)) {
d2 = curcost + data[curv][nv].p;
}
int d3 = min(d1, d2);
int npath = curpath | (1 << nv);
if (d3 < dist[nv][npath]) {
dist[nv][npath] = d3;
q.push(make_pair(make_pair(nv, npath), d3));
}
}
}
int res = INF;
for (int i = 0; i < (1 << 11); i++) {
res = min(res, dist[N][i]);
}
return res;
}
int main() {
scanf("%d %d", &N, &M);
for (int i = 0; i < M; i++) {
int a, b, c, p, r;
scanf("%d %d %d %d %d", &a, &b, &c, &p, &r);
g[a].push_back(b);
data[a][b] = (struct node) {c, p, r};
}
int ret = bfs();
if (ret == INF) printf("impossible\n");
else printf("%d\n", ret);
return 0;
}
我认为问题可能在于您的 data[a][b]
结构假定从 a 到 b 最多只有一条路。
但是,问题是:
There may be more than one road connecting one city with another.
问题link:http://www.spoj.com/problems/MMINPAID/
提交时获得 WA。 我使用 bfs 到达第 N 个节点并使用位掩码计算路径中节点的最小成本。然而,在 运行 大量测试用例并与公认的解决方案进行比较之后,无法找到失败的测试用例。 代码:
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int MAXN = 15, INF = 1 << 29;
struct node {
int c, p, r;
};
struct node data[MAXN][MAXN];
vector<int> g[MAXN];
int N, M, dist[MAXN][1 << 11];
int bfs() {
for (int i = 0; i < MAXN; i++)
for (int k = 0; k < (1 << 11); k++)
dist[i][k] = INF;
queue< pair< pair <int, int> , int > > q;
int v = 1, path = (1 << 1), cost = 0;
dist[v][path] = 0;
q.push(make_pair(make_pair(v, path), cost));
while (!q.empty()) {
int curv = q.front().first.first;
int curpath = q.front().first.second;
int curcost = q.front().second;
q.pop();
for (int i = 0; i < g[curv].size(); i++) {
int nv = g[curv][i];
int d1 = curcost + data[curv][nv].r;
int d2 = INF;
if (curpath & (1 << data[curv][nv].c)) {
d2 = curcost + data[curv][nv].p;
}
int d3 = min(d1, d2);
int npath = curpath | (1 << nv);
if (d3 < dist[nv][npath]) {
dist[nv][npath] = d3;
q.push(make_pair(make_pair(nv, npath), d3));
}
}
}
int res = INF;
for (int i = 0; i < (1 << 11); i++) {
res = min(res, dist[N][i]);
}
return res;
}
int main() {
scanf("%d %d", &N, &M);
for (int i = 0; i < M; i++) {
int a, b, c, p, r;
scanf("%d %d %d %d %d", &a, &b, &c, &p, &r);
g[a].push_back(b);
data[a][b] = (struct node) {c, p, r};
}
int ret = bfs();
if (ret == INF) printf("impossible\n");
else printf("%d\n", ret);
return 0;
}
我认为问题可能在于您的 data[a][b]
结构假定从 a 到 b 最多只有一条路。
但是,问题是:
There may be more than one road connecting one city with another.