POJ ~ 1502 ~ MPI Maelstrom （Dijkstra + 字符处理）

题意：N个点。下面N-1行，给出你一些无向边，‘x’表示不互通。问1点到哪个点的距离最远，输出距离？

解释下样例，

5

50

30 5

100 20 50

10 x x 10

有五个点1~5，1到1的距离为0，没有给出。第二行表示，2到1距离为50。第三行表示3到1距离为30，3到2距离为5。第四行表示4到1距离为100，4到2距离为20，4到3距离为50。第五行表示5到1距离为10，5和2不互通，5和3不互通，5到4距离为100。

思路：输入的时候，输入字符串。如果为x就表示不互通，不加边。如果不是x，就用stringstream把字符串变为数字，加双向边。跑一边Dijkstra，求一个最远值。

优先队列优化Dijkstra:

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<sstream>
#include<queue>
#include<vector>
using namespace std;
const int MAXN = 1e5 + 6;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to, dist; //起点,终点,距离
Edge(int u, int v, int w):from(u), to(v), dist(w) {}
};

struct Dijkstra
{
int n, m; //结点数,边数(包括反向弧)
vector<Edge> edges; //边表。edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
int vis[MAXN]; //标记数组
int d[MAXN]; //s到各个点的最短路
int pre[MAXN]; //上一条弧

void init(int n)
{
this->n = n;
edges.clear();
for (int i = 0; i <= n; i++) G[i].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);
}

struct HeapNode
{
int u, d;
bool operator < (const HeapNode& rhs) const
{
return d > rhs.d;
}
HeapNode(int from, int w): u(from), d(w) {}
};

void dijkstra(int s)
{
priority_queue<HeapNode> Q;
for (int i = 0; i <= n; i++) d[i] = INF;
d[s] = 0;
memset(vis, 0, sizeof(vis));
Q.push(HeapNode(s, 0));
while (!Q.empty())
{
HeapNode x = Q.top(); Q.pop();
int u = x.u;
if (vis[u]) continue;
vis[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)
{
d[e.to] = d[u] + e.dist;
pre[e.to] = G[u][i];
Q.push(HeapNode(e.to, d[e.to]));
}
}
}
}
};
Dijkstra solve;
int main()
{
int n;
while (~scanf("%d", &n))
{
solve.init(n);
for (int i = 2; i <= n; i++)
{
for (int j = 1; j < i; j++)
{
int w;
string s; cin >> s;
if (s != "x")

e314
{
stringstream ss(s);
ss >> w;
solve.add_edge(i, j, w);
solve.add_edge(j, i, w);
}
}
}
solve.dijkstra(1);
int MAX = -INF;
for (int i = 2; i <= n; i++)
{
MAX = max(MAX, solve.d[i]);
}
printf("%d\n", MAX);
}
return 0;
}
/*
5
50
30 5
100 20 50
10 x x 10
*/


Dijkstra：

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<sstream>
#include<queue>
#include<vector>
using namespace std;
const int MAXN = 1e5 + 6;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to, dist;       //起点,终点,距离
Edge(int u, int v, int w):from(u), to(v), dist(w) {}
};

struct Dijkstra
{
int n, m;                 //结点数,边数(包括反向弧)
vector<Edge> edges;       //边表。edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN];      //邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
int vis[MAXN];            //标记数组
int d[MAXN];              //s到各个点的最短路
int pre[MAXN];            //上一条弧

void init(int n)
{
this->n = n;
edges.clear();
for (int i = 0; i <= n; i++) G[i].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)
{
for (int i = 0; i <= n; i++) d[i] = INF;
d[s] = 0;
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= n; i++)
{
int  pos, MIN = INF;
for (int j = 1; j <= n; j++)
{
if(!vis[j] && d[j] <= MIN) MIN = d[pos = j];
}
vis[pos] = true;
for (int j = 0; j < G[pos].size(); j++)
{
Edge& e = edges[G[pos][j]];
if (d[e.to] > d[pos] + e.dist)
{
d[e.to] = d[pos] + e.dist;
pre[e.to] = G[pos][j];
}
}
}
}
};
Dijkstra solve;
int main()
{
int n;
while (~scanf("%d", &n))
{
solve.init(n);
for (int i = 2; i <= n; i++)
{
for (int j = 1; j < i; j++)
{
int w;
string s; cin >> s;
if (s != "x")
{
stringstream ss(s);
ss >> w;
solve.add_edge(i, j, w);
solve.add_edge(j, i, w);
}
}
}
solve.dijkstra(1);
int MAX = -INF;
for (int i = 2; i <= n; i++)
{
MAX = max(MAX, solve.d[i]);
}
printf("%d\n", MAX);
}
return 0;
}
/*
5
50
30 5
100 20 50
10 x x 10
*/