題意分析:
區間[ai, bi]至少有ci個點在集合Z中,現在問:集合Z中,元素最少是多少個?
解題思路:
題目相當於是說點[ai,bi + 1)間,有>=ci的點在集合Z中,即:ai->bi+1 >= ci。把點看作圖中的點,則有:dis[bi + 1] >= ci + dis[ai],翻譯成轉移語言:if (dis[bi + 1] < ci + dis[ai]) dis[bi + 1] = dis[ai] + ci。就是求圖中的最長距離了。這樣還遺留一個問題,起始點到終點路不全啊,這個時候我們就要挖掘隱含條件了1 >= dis[i + 1] - dis[i] >= 0。這樣就能建圖了。答案就是從最小點到最大點的最長路徑。
個人感受:
老是不能很好地根據不等式確定是怎麼樣的求路徑。用轉移寫出來就好理解太多了,贊!另外隱含條件沒好好去想,列出式子就GG了。。。。
具體代碼如下:
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<string>
#define pr(x) cout << #x << " = " << (x) << '\n';
using namespace std;
const int INF = 0x7f7f7f7f;
const int MAXN = 5e4 + 111;
const int MAXM = 2e5 + 111;
struct Edge {
int to, next, w;
}edge[MAXM];
int head[MAXN], tol, dis[MAXN];
void addedge(int u, int v, int w) {
edge[tol].to = v;
edge[tol].next = head[u];
edge[tol].w = w;
head[u] = tol++;
}
bool in[MAXN];
void spfa(int s) {
queue<int> q;
q.push(s);
dis[s] = 0;
in[s] = 1;
while (q.size()) {
int u = q.front(); q.pop();
in[u] = 0;
for (int i = head[u]; ~i; i = edge[i].next) {
int v = edge[i].to;
if (dis[v] < dis[u] + edge[i].w) {
dis[v] = dis[u] + edge[i].w;
if (!in[v]) {
in[v] = 1;
q.push(v);
}
}
}
}
}
int main()
{
int n, mi = INF, mx = 0, u, v, w;
memset(head, -1, sizeof head);
tol = 0;
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d%d%d", &u, &v, &w);
addedge(u, v + 1, w);
mi = min(mi, u);
mx = max(mx, v + 1);
}
for (int i = mi; i < mx; ++i) {
addedge(i, i + 1, 0);
addedge(i + 1, i, -1);
dis[i] = -INF;
}
dis[mx] = -INF;
spfa(mi);
printf("%d\n", dis[mx]);
return 0;
}