CodeForces CF 360 E Levko and Game貪欲+SPFA
5548 ワード
欲張りな問題だそうです.の
N個の点が知られており、M+Kバーにはエッジとエッジの重みがあり、重いエッジと自己リングがある可能性がある.
所定のk辺の一部の辺の辺権を変更するには、費用を消費することができる.
第i条エッジ重み修正範囲[li,ri].
与えられたK辺の長さをs 1からfまでs 2からfまでの時間より短くする.
エッジの場合、dis(s 1,x)のエッジ長をlに変更することで、dis(s 1,f)をより小さくすることができる.
このように修正+再走最短路後dis(s 1,f)引き分けについては,dis(s 1,x)<=dis(s 2,x)と判断すればよい.
必ずif(!check(0))if(!check(1))を書くことができないことに気づいた&....条件を満たさずに実行しないと約束した最適化ですね.の
Levko loves sports pathfinding competitions in his city very much. In order to boost his performance, Levko spends his spare time practicing. The practice is a game.
The city consists of n intersections connected by m + k directed roads. Two or more roads can connect the same pair of intersections. Besides, there can be roads leading from an intersection to itself.
Levko and Zenyk are playing a game. First Levko stands on intersection s1, and Zenyk stands on intersection s2. They both want to get to intersection f. The person who does it quicker wins. If they get there at the same time, the game ends with a draw. By agreement both players start simultaneously and move with the same speed.
Levko wants to win very much. He knows the lengths of all the roads in the city. Also he knows that he can change the lengths of some roads (there are k such roads at all) if he pays the government. So, the government can change the length of the i-th road to any integer value in the segment [li, ri] (both borders inclusive). Levko wondered if he can reconstruct the roads so as to win the game and whether he can hope for the draw if he cannot win.
You should consider that both players play optimally well. It is guaranteed that we can get from intersections s1 and s2 to intersection f.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 104, 1 ≤ k ≤ 100). The second line contains three integers s1, s2 and f(1 ≤ s1, s2, f ≤ n).
The next m lines contains the descriptions of the roads that cannot be changed by Levko. Each line contains three integers ai, bi and ci(1 ≤ ai, bi ≤ n, 1 ≤ ci ≤ 109), representing a road from intersection ai to intersection bi of length ci.
The next k lines contains the descriptions of the roads that can be changed by Levko. Each line contains four integers ai, bi, li and ri(1 ≤ ai, bi ≤ n, 1 ≤ li ≤ ri ≤ 109), representing a road from intersection ai to intersection bi, Levko can set the road's length within limits[li, ri].
Consider all intersections numbered from 1 to n. It is guaranteed that you can get from intersections s1 and s2 to intersection f.
Output
In the first line print string "WIN"(without the quotes) if Levko can win this game, string "DRAW"(without the quotes) if Levko can end the game with a draw and "LOSE"(without the quotes) if he loses for sure.
If the answer is "WIN"or "DRAW", then print on the second line k space-separated integers — the length of the roads Levko sets in the order they occur in the input.
Sample test(s)
input
output
input
output
input
output
N個の点が知られており、M+Kバーにはエッジとエッジの重みがあり、重いエッジと自己リングがある可能性がある.
所定のk辺の一部の辺の辺権を変更するには、費用を消費することができる.
第i条エッジ重み修正範囲[li,ri].
与えられたK辺の長さをs 1からfまでs 2からfまでの時間より短くする.
エッジ
このように修正+再走最短路後dis(s 1,f)
必ずif(!check(0))if(!check(1))を書くことができないことに気づいた&....条件を満たさずに実行しないと約束した最適化ですね.の
#include <cstdio>
#include <cstring>
#define FOR(i,j,k) for(i=j;i<=k;i++)
int read() {
int s=0,f=1;char ch=getchar();
for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
for(;'0'<=ch&&ch<='9';ch=getchar())s=s*10+ch-'0';
return s*f;
}
typedef long long ll;
const int N = 10001, M = 20001;
ll da[N], db[N], el[M], er[M];
int n, m, k, s1, s2, t;
int head[N], next[M], from[M], to[M], q[N * 64], cnt = 0;
bool vis[N];
void add(int u, int v, int l, int r) {
next[++cnt] = head[u]; from[cnt] = u; to[cnt] = v;
el[cnt] = l; er[cnt] = r; head[u] = cnt;
}
void spfa(ll dis[N], int s) {
memset(dis, 0x7f, sizeof(ll)*N);
memset(vis, 0, sizeof vis);
int f = 0, r = 0, i, u;
q[r++] = s; dis[s] = 0; vis[s] = 1;
while (f < r) {
u = q[f++];
for (i = head[u]; i; i = next[i]) {
if (dis[to[i]] > dis[u] + er[i]) {
dis[to[i]] = dis[u] + er[i];
if (!vis[to[i]]) {
q[r++] = to[i];
vis[to[i]] = 1;
}
}
}
vis[u] = 0;
}
}
bool check(int f) {
int flag = 1, i;
while (flag) {
flag = 0;
spfa(da, s1); spfa(db, s2);
FOR(i,m+1,m+k)
if(da[from[i]]<db[from[i]] + f && el[i] != er[i])
er[i] = el[i], flag = 1;
if (da[t] < db[t] + f) {
puts(f ? "DRAW" : "WIN");
FOR(i,m+1,m+k) printf("%d ", er[i]);
return 1;
}
}
return 0;
}
int main() {
n = read(), m = read(), k = read();
s1 = read(), s2 = read(), t = read();
int i, x, y, z, l, r;
FOR(i,1,m) x=read(),y=read(),z=read(),add(x,y,z,z);
FOR(i,1,k) x=read(),y=read(),l=read(),r=read(),add(x,y,l,r);
if(!check(0)) if(!check(1)) puts("LOSE");
return 0;
}
Levko loves sports pathfinding competitions in his city very much. In order to boost his performance, Levko spends his spare time practicing. The practice is a game.
The city consists of n intersections connected by m + k directed roads. Two or more roads can connect the same pair of intersections. Besides, there can be roads leading from an intersection to itself.
Levko and Zenyk are playing a game. First Levko stands on intersection s1, and Zenyk stands on intersection s2. They both want to get to intersection f. The person who does it quicker wins. If they get there at the same time, the game ends with a draw. By agreement both players start simultaneously and move with the same speed.
Levko wants to win very much. He knows the lengths of all the roads in the city. Also he knows that he can change the lengths of some roads (there are k such roads at all) if he pays the government. So, the government can change the length of the i-th road to any integer value in the segment [li, ri] (both borders inclusive). Levko wondered if he can reconstruct the roads so as to win the game and whether he can hope for the draw if he cannot win.
You should consider that both players play optimally well. It is guaranteed that we can get from intersections s1 and s2 to intersection f.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 104, 1 ≤ k ≤ 100). The second line contains three integers s1, s2 and f(1 ≤ s1, s2, f ≤ n).
The next m lines contains the descriptions of the roads that cannot be changed by Levko. Each line contains three integers ai, bi and ci(1 ≤ ai, bi ≤ n, 1 ≤ ci ≤ 109), representing a road from intersection ai to intersection bi of length ci.
The next k lines contains the descriptions of the roads that can be changed by Levko. Each line contains four integers ai, bi, li and ri(1 ≤ ai, bi ≤ n, 1 ≤ li ≤ ri ≤ 109), representing a road from intersection ai to intersection bi, Levko can set the road's length within limits[li, ri].
Consider all intersections numbered from 1 to n. It is guaranteed that you can get from intersections s1 and s2 to intersection f.
Output
In the first line print string "WIN"(without the quotes) if Levko can win this game, string "DRAW"(without the quotes) if Levko can end the game with a draw and "LOSE"(without the quotes) if he loses for sure.
If the answer is "WIN"or "DRAW", then print on the second line k space-separated integers — the length of the roads Levko sets in the order they occur in the input.
Sample test(s)
input
4 1 3
1 3 4
3 2 2
1 2 1 3
2 4 1 3
3 4 1 3
output
WIN
1 1 3 input
4 1 3
1 3 4
3 2 2
1 2 1 3
2 4 1 3
3 4 1 2
output
DRAW
1 1 2 input
5 4 2
1 2 5
1 3 3
1 4 4
2 3 2
2 4 3
3 5 1 5
4 5 4 7
output
LOSE