2 2 2 1 2 13 2 1 33 4 6 1 2 10 2 1 60 1 3 20 3 4 10 2 4 5 4 1 50
46 210
题目分析：这题HDU的数据比较水，给的时间也短，POJ的数据超int，wa了几发，做法就是1到各点直接spfa搞一下，反过来，就把图也反过来建一下，再搞一次spfa就可以了，一套好板子也是很重要的，书上板子动态分配伤不起，其实vector也挺费时，在poj跑了4000ms+，但是比较好写
POJ:
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#define ll long long
int const MAX = 1e6 + 5;
ll const INF = 0xffffffff;
using namespace std;
struct NODE
{
int v, w;
NODE(int vv, ll ww)
{
v = vv;
w = ww;
}
};
struct EDGE
{
int u, v;
ll w;
}e[MAX];
vector <NODE> vt[MAX];
int p, q;
ll ans, dist[MAX];
bool vis[MAX];
void SPFA(int v0)
{
memset(vis, false, sizeof(vis));
for(int i = 0; i <= p; i++)
dist[i] = INF;
dist[v0] = 0;
queue <int> Q;
Q.push(v0);
while(!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = false;
int sz = vt[u].size();
for(int i = 0; i < sz; i++)
{
int v = vt[u][i].v;
ll w = vt[u][i].w;
if(dist[v] > dist[u] + w)
{
dist[v] = dist[u] + w;
if(!vis[v])
{
Q.push(v);
vis[v] = true;
}
}
}
}
}
void Clear()
{
for(int i = 1; i <= p; i++)
vt[i].clear();
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
ans = 0;
scanf("%d %d", &p, &q);
for(int i = 0; i < q; i++)
scanf("%d %d %lld", &e[i].u, &e[i].v, &e[i].w);
Clear();
for(int i = 0; i < q; i++)
vt[e[i].u].push_back(NODE(e[i].v, e[i].w));
SPFA(1);
for(int i = 1; i <= p; i++)
ans += dist[i];
Clear();
for(int i = 0; i < q; i++)
vt[e[i].v].push_back(NODE(e[i].u, e[i].w));
SPFA(1);
for(int i = 1; i <= p; i++)
ans += dist[i];
printf("%lld\n", ans);
}
}HDU:
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
int const MAX = 1e6 + 5;
int const INF = 0xfffffff;
using namespace std;
struct NODE
{
int v, w;
NODE(int vv, int ww)
{
v = vv;
w = ww;
}
};
struct EDGE
{
int u, v, w;
}e[MAX];
vector <NODE> vt[MAX];
int p, q, ans, dist[MAX];
bool vis[MAX];
void SPFA(int v0)
{
memset(vis, false, sizeof(vis));
for(int i = 0; i <= p; i++)
dist[i] = INF;
dist[v0] = 0;
queue <int> Q;
Q.push(v0);
while(!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = false;
int sz = vt[u].size();
for(int i = 0; i < sz; i++)
{
int v = vt[u][i].v;
int w = vt[u][i].w;
if(dist[v] > dist[u] + w)
{
dist[v] = dist[u] + w;
if(!vis[v])
{
Q.push(v);
vis[v] = true;
}
}
}
}
}
void Clear()
{
for(int i = 1; i <= p; i++)
vt[i].clear();
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
ans = 0;
scanf("%d %d", &p, &q);
for(int i = 0; i < q; i++)
scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
Clear();
for(int i = 0; i < q; i++)
vt[e[i].u].push_back(NODE(e[i].v, e[i].w));
SPFA(1);
for(int i = 1; i <= p; i++)
ans += dist[i];
Clear();
for(int i = 0; i < q; i++)
vt[e[i].v].push_back(NODE(e[i].u, e[i].w));
SPFA(1);
for(int i = 1; i <= p; i++)
ans += dist[i];
printf("%d\n", ans);
}
}HDU 1535 && POJ 1511 Invitation Cards (SPFA 模板 ＋ 反向建图)
原文地址：http://blog.csdn.net/tc_to_top/article/details/43676145
