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UVa 10917 A Walk through the forest

2016-07-21 19:33 274 查看
题目大意:

给你n个点m条边的无向图, 问从起点(1号点)到达终点(2号点)的总路径数量(要求从每个点u走到下一个点v时满足:v到终点的距离小于u到终点的距离。

分析:

从起点出发到每个点的最短路径可以用一次Dijkstra算法求出,那么原问题就转化成了在一个Dag上求路径数目,可以直接DP或者利用Dijkstra计算出的虽短路树来求解

代码:

#include<bits/stdc++.h>
using namespace std;
const int inf = 0x7fffffff;
const int maxn = 1000 + 10;

struct edge {
int from, to, dist;
};

struct Heapnode {
int d, u;
bool operator < (const Heapnode& rhs) const {
return d > rhs.d;
}
};

struct Dijkstra {
int n, m;
vector<edge> edges;
vector<int> G[maxn];
bool done[maxn];
int dis[maxn], pre[maxn];

void init(int n) {
this->n = n;
for(int i=0; i<n; i++) G[i].clear();
edges.clear();
}
void add_edge(int from, int to, int dist) {
edges.push_back((edge){from, to, dist});
m = edges.size();
G[from].push_back(m-1);
}
void dijkstra(int s) {
priority_queue<Heapnode> q;
memset(pre, 0, sizeof(pre));
memset(done, false, sizeof(done));
for(int i=0; i<n; i++) dis[i] = inf; dis[s] = 0;

q.push((Heapnode){0, s});
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<(int)G[u].size(); i++) {
edge& e = edges[G[u][i]];
if(dis[e.to] > dis[u] + e.dist) {
dis[e.to] = dis[u] + e.dist;
pre[e.to] = G[u][i];
q.push((Heapnode){dis[e.to], e.to});
}
}
}
}
}Dij;

int n, m, u, v, w;
int ways[maxn];

int dfs(int u) {
if(u == 1) return 1;
int& ans = ways[u];
if(ans >= 0) return ans;

ans = 0;
for(int i=0; i<(int)Dij.G[u].size(); i++) {
int v = Dij.edges[Dij.G[u][i]].to;
if(Dij.dis[u] > Dij.dis[v]) ans += dfs(v);
}
return ans;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("data.txt", "r", stdin);
freopen("ans.txt", "w", stdout);
#endif
while(scanf("%d%d", &n, &m) == 2 && n+m) {
Dij.init(n);
for(int i=0; i<m; i++) {
scanf("%d%d%d", &u, &v, &w); u--, v--;
Dij.add_edge(u, v, w);
Dij.add_edge(v, u, w);
}
Dij.dijkstra(1);
memset(ways, -1, sizeof(ways));
printf("%d\n", dfs(0));
}
return 0;
}
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标签:  uva dijkstra 算法 dp