POJ_1062_最短路变形

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POJ 1062

题意：

替换就相当于一条指向另一个点的有向边,权值是价值,但是每个点都有等级作为限制条件,
限制条件还必须在一个范围,例如:
level[1] = 3 m=1,节点等级限制范围:[2,3]、[3,4],
level[1] = 3 m=2, 范围：[1,3]、[2,4],[3,5]
3.每次最短路在将不再范围的点排除,然后取每个范围中最短的点.

复制代码

#include <iostream>
#include <queue>
#include <string.h>
using namespace std;
const int maxn = 105;
const int inf = 0x3f3f3f3f;
int e[maxn][maxn];
int dist[maxn];
int level[maxn];
bool vis[maxn];
int value[maxn];
int n,radium;
int dij() {
priority_queue<pair<int,int> > que;
memset(dist,0x3f,sizeof dist);
dist[1] = 0;
que.push( make_pair(-dist[1],1) );
while (!que.empty()) {
int cur = que.top().second; que.pop();
if (vis[cur]) continue;
vis[cur] = 1;
for (int i=1; i<=n; i++) {
if (!vis[i] && dist[i] > dist[cur] + e[cur][i]) {
dist[i] = dist[cur] + e[cur][i];
que.push( make_pair(-dist[i],i) );
}
}
}
int MIN = inf;
for (int i=1; i<=n; i++)
MIN = min(MIN,dist[i]+value[i]);
return MIN;
}
int main() {
//	freopen("a.txt","r",stdin);
cin >> radium >> n;
memset(e,0x3f,sizeof e);
int k;
for (int i=1; i<=n; i++) {
cin >> value[i] >> level[i] >> k;
for (int j=1; j<=k; j++) {
int to,div;
cin >> to >> div;
e[i][to] = div;
}
}
int ans = inf;
for (int i=0; i<=radium; i++) { // 枚举一路的等级区间范围
memset(vis,0,sizeof vis);
for (int j=1; j<=n; j++)
// 把超出等级范围的点这一轮排除掉
if (level[1] - radium + i > level[j] || level[j] > level[1] + i)
vis[j] = 1;
ans = min(ans,dij());
}
cout << ans;
return 0;
}

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#include <iostream>
#include <queue>
#include <string.h>
using namespace std;
const int maxn = 105;
const int inf = 0x3f3f3f3f;
int e[maxn][maxn];
int dist[maxn];
int level[maxn];
bool vis[maxn];
int value[maxn];
int n,radium;
int dij() {
priority_queue<pair<int,int> > que;
memset(dist,0x3f,sizeof dist);
dist[1] = 0;
que.push( make_pair(-dist[1],1) );
while (!que.empty()) {
int cur = que.top().second; que.pop();
if (vis[cur]) continue;
vis[cur] = 1;
for (int i=1; i<=n; i++) {
if (!vis[i] && dist[i] > dist[cur] + e[cur][i]) {
dist[i] = dist[cur] + e[cur][i];
que.push( make_pair(-dist[i],i) );
}
}
}
int MIN = inf;
for (int i=1; i<=n; i++)
MIN = min(MIN,dist[i]+value[i]);
return MIN;
}
int main() {
//	freopen("a.txt","r",stdin);
cin >> radium >> n;
memset(e,0x3f,sizeof e);
int k;
for (int i=1; i<=n; i++) {
cin >> value[i] >> level[i] >> k;
for (int j=1; j<=k; j++) {
int to,div;
cin >> to >> div;
e[i][to] = div;
}
}
int ans = inf;
for (int i=0; i<=radium; i++) { // 枚举一路的等级区间范围
memset(vis,0,sizeof vis);
for (int j=1; j<=n; j++)
// 把超出等级范围的点这一轮排除掉
if (level[1] - radium + i > level[j] || level[j] > level[1] + i)
vis[j] = 1;
ans = min(ans,dij());
}
cout << ans;
return 0;
}